Please discuss transformer losses vs %Z

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I know what transformer impedance is and how it is calculated. It is, of course, a somewhat unique use of the word "impedance" How does %Z relate to load and no load losses? I suspect there are other factors at play where we could have transformer A that has a lower %Z yet higher no load losses than transformer B with a higher %Z. In fact I know this is true from measurements I have made. I dont quite understand this. Anyone want to take a stab at this?
 
I know what transformer impedance is and how it is calculated. It is, of course, a somewhat unique use of the word "impedance" How does %Z relate to load and no load losses? I suspect there are other factors at play where we could have transformer A that has a lower %Z yet higher no load losses than transformer B with a higher %Z. In fact I know this is true from measurements I have made. I dont quite understand this. Anyone want to take a stab at this?

It’s about %Z, X/R ratio and MVA base (which essentially boil down to resistive elements in a Steinmetz approximate model). The %Z is just a percentage of base impedance, so you cannot compare transformer A and transformer B unless they are compared at the same base power. Transformer losses are a function of core loss (hysteresis/excitation) and copper loss.

When determining no load and full load losses, the Steinmetz model is reduced to a simple series circuit with an equivalent %Z.


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The %Z given on a transformer nameplate does take into account the all the losses however as Xptpcrewx points out you must convert the %Z values to the actual impedance. For any transformer the nameplate %Z is given assuming the base kVA or MVA is the rating of the transformer. So if you have a 100 kVA transformer, the base kVA is 100. The base voltage is the nameplate voltage. If you want to know the actual impedance referenced to the primary side, you would use the primary nameplate voltage. IF you want the impedance referenced from the secondary side of the transformer you would use the nameplate secondary voltage. As an example, I assume a 100 kVA transformer with a secondary side voltage of 240/120 volts and a %Z = 2.5. To find the actual impedance, you need to multiply the the (%Z/100) by the based impedance which is defined as (base voltage squared)/base kVA. In this case the base impedance is (240*240)/100 kVA = 57,600/100,000=0.576, Z-actual then is = (2.5/100)*(0.579)= 0.0144 ohms. What you do not have however is the X/R ratio for your specific transformer but these value are published by the manufacturer or average values are used by the IEEE. The range for this size transformer is between about 1.5 to 2.7 for the X/R ratio. I will use 1.5. Tan^-1 (1.5) = 56.3° so the impedance of this transformer is 0.0144 Ohms at 56.3° will give you R = 0.008, Xl = 0.012. Hope this helps.
 
The %Z given on a transformer nameplate does take into account the all the losses however as Xptpcrewx points out you must convert the %Z values to the actual impedance. For any transformer the nameplate %Z is given assuming the base kVA or MVA is the rating of the transformer. So if you have a 100 kVA transformer, the base kVA is 100. The base voltage is the nameplate voltage. If you want to know the actual impedance referenced to the primary side, you would use the primary nameplate voltage. IF you want the impedance referenced from the secondary side of the transformer you would use the nameplate secondary voltage. As an example, I assume a 100 kVA transformer with a secondary side voltage of 240/120 volts and a %Z = 2.5. To find the actual impedance, you need to multiply the the (%Z/100) by the based impedance which is defined as (base voltage squared)/base kVA. In this case the base impedance is (240*240)/100 kVA = 57,600/100,000=0.576, Z-actual then is = (2.5/100)*(0.579)= 0.0144 ohms. What you do not have however is the X/R ratio for your specific transformer but these value are published by the manufacturer or average values are used by the IEEE. The range for this size transformer is between about 1.5 to 2.7 for the X/R ratio. I will use 1.5. Tan^-1 (1.5) = 56.3° so the impedance of this transformer is 0.0144 Ohms at 56.3° will give you R = 0.008, Xl = 0.012. Hope this helps.
What Im not clear on here is the part in red. Can compute no load losses from here, or is that not where you are going? IF not, can we calculate no load losses (given %Z and X/R) or does that just have to be measured?


So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?
 
So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?

Put another way,:say someone puts a bunch of 12.47GY/7200-120/240 25KVA transformers in front of you. the impedances range from 1.6 to 2.4. %Z They put a gun to your head and tell you to pick the one with the least no load losses and/or load losses. What is the likelihood that picking the one with the lowest %Z will keep you from getting a bullet in the head?
 
What Im not clear on here is the part in red. Can compute no load losses from here, or is that not where you are going? IF not, can we calculate no load losses (given %Z and X/R) or does that just have to be measured?


So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?

The losses are usually determined at the factory from testing. They can also be derived from open and short-circuit tests. Some large power transformers put loss/efficiency information directly on the nameplate, some smaller distribution type transformers have it published in their data sheet. %Z by itself provides no indication of the transformer no load or full load losses. In general however, lower impedance results in lower losses. As mentioned %Z is just a percentage. Percentages without a reference quantity are meaningless.


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Put another way,:say someone puts a bunch of 12.47GY/7200-120/240 25KVA transformers in front of you. the impedances range from 1.6 to 2.4. %Z They put a gun to your head and tell you to pick the one with the least no load losses and/or load losses. What is the likelihood that picking the one with the lowest %Z will keep you from getting a bullet in the head?

If the transformers are all rated the same with %Z being the only difference between the units, then the lower the %Z, the lower the losses.


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How does %Z relate to load and no load losses? I suspect there are other factors at play where we could have transformer A that has a lower %Z yet higher no load losses than transformer B with a higher %Z.
A significant part of the %Z is due to leakage reactance, which results from that part of the magnetic flux produced by the primary winding that does not pass through the secondary winding. Transformers used in welders, neon sign transformers, and other applications which need a high impedance for current limiting purposes are built so they have a high leakage reactance. But that doesn't necessarily imply that they have high losses.
Leakage inductance is the parameter that can most easily be adjusted by a transformer manufacturer to control the %Z.

As Xptpcrewx mentioned, no load losses are caused by hysteresis losses (from overcoming the remnant magnetization left from the previous half-cycle) and from resistive losses in the core caused by induced eddy currents There are copper losses under no-load, but they would typically be relatively small with only the excitation current flowing.
 
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In general however, lower impedance results in lower losses. As mentioned %Z is just a percentage. Percentages without a reference quantity are meaningless.


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If the transformers are all rated the same with %Z being the only difference between the units, then the lower the %Z, the lower the losses.


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I should have made clear that when trying to gauge losses but looking at %Z, I was assuming each would have the same KVA.
 
What Im not clear on here is the part in red. Can compute no load losses from here, or is that not where you are going? IF not, can we calculate no load losses (given %Z and X/R) or does that just have to be measured?


So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?
The value you would calculate for the actual impedance includes all losses. If you wanted to know what portion makes up each part you must test the transformer to measure the no load losses (transformer energized and operating open circuit on the load side of the transformer), this would give the losses due to hysteresis and eddy currents. Then perform a short circuit test to measure the copper losses.
As to the part in red, I found the magnitude of the transformer impedance in actual ohms as 0.0144 ohms. The impedance angle of 56.3 degrees comes from the X/R ratio of 1.5. If you look at an impedance triangle you will find the R value on the x-axis and perpendicular to this you will find the X (reactive componen) on the y-axis. You may determine Z = square root of [(R*R)+(X*X)]. The ratios of the various sides of a right triangle have been labeled with names such Tangent, Sine, Cosine of the angles made by the construction of these triangles. The X/R ratio is the Tangent of an angle that has the exact ratio of X/R. If we know the ratio of the y-axis (X) / x-axis (R) is the Tangent, the ArcTangent of (X/R) will give us the impedance angle. To find the value of R = Z*Cosine (56.3), X = Z*Sine (56.3). Most calculators can do this for you by using the Sin, Cos and Tan functions or their reverse operations usually shown as Sin to the -1, or Cos to the -1, or Tan to the -1, also called the Arcsin, Arccos or Arctan. Sorry for the long answer. I hope it helps.
 
Ok I took some no load loss measurements today on three 4160GY/2400 - 120/240 15 KVA pads (actually one is 240 only, no center tap), and one 2400 - 12/240 dry type. Watt measurements were taken with two different meters, an amprobe ACDC54NAV and a EKM omnimeter. I trust the EKM a bit more for these low watt measurements. The EKM value was always the higher value and are presented in Amprobe/EKM format. The only disclaimer here is that I dont have PT's and these are in operation, so for convenience, some of these measurements are taken indirectly by subtracting the value of the first unit. I am assuming that doesnt add much inaccuracy. Maybe I am wrong?

T1 is an older reman unit. It does not have a date of manufacturer. It appears to made by Howard. %Z is 1.98. Watts= 117/116 (measured thru T4)
T2 is a cooper, date of manufacture 2010, meets DOE 2010, %Z is 1.9. Watts=35/40 (measured directly)
T3 is a cooper, date of manufacture 2009, no DOE on label (Im not sure what was before DOE 2010), %Z is 1.7. Watts=55/56 (measured thru T4)
T4 is the dry type, not sure of date of manufacture or impedance. Watts= 120/136 (measured directly

Granted this is a small sample, but the newer units seem to have much lower losses. I browsed thru inventory of some of these transformer re-manufactures (such as T&R electric) and they never seem to put the mfg date on the new label. Now I think it is safe to assume most of these are "oldish" and would have higher losses, yet quite a few of them have impedance that is quite low. Granted I am making some assumptions here, but it seems like %Z doesnt correlate well to no load losses (my data seems to show this as well).
 

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some of these measurements are taken indirectly by subtracting the value of the first unit
Can you elaborate how you measured the voltage? I don’t understand what you mean here. Thanks.


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One thing I should clarify here. You are taking no load measurements which are only core losses (excitation/hysteresis). %Z doesn’t have anything to do with this. %Z is the effective impedance “through” the transformer. If you want to correlate %Z to losses, it needs to be for copper losses (as done with short-circuit tests).

I commend you on your curiosity to carry out the experiment but I think you could have gotten this information from the manufacturer published no load efficiency.

The reason you are seeing higher no load losses with older units is because older units were probably not constructed with the same energy efficient requirements as newer units and are hence constructed with higher loss core materials.


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So T4 is a step up unit. T1 and T3 are step down. I measured the no load loss of T4 directly, nothing connected to it. Then I re-connected T1, and measured the primary side of T4 again, subtracted the T4 loss figure, and assumed that is the T1 Loss figure. Did the same for T3.
 
Xptpcrewx is correct in that the open circuit test will only measure the losses due to hysteresis and eddy currents. Further, these losses only occur due to the current flow in the primary winding (primary in this sense means the source side of the transformer). The losses measured under the no-load condition are ignored for the most part to simplify calculations based on the very small loss of accuracy of the final result. It appears from your measurements you are attempting determine the %Z using the open circuit type of test. The percent Z represents a measurement of the equivalent resistance and reactance values of the transformer. That is it takes into account the primary and secondary resistance and reactance to give you an equivalent series impedance that you may use to determine the voltage regulation (voltage drop) that will occur based the load current and power factor.

An open circuit test will give you the no-load losses that are used with the short circuit test for the load losses, to calculate the overall efficiency of the transformer. I note here that the maximum efficiency of any transformer occurs when the open circuit test losses and the copper losses are equal, typically around 60% of the full load capacity.

To equate the %Z on the nameplate of the transformer to the actual equivalent %R and %X, you need to perform a short circuit test. The easiest path to this number is to determine the short circuit power (Psc), derive Re (equivalent Resistance) = Psc/(Isc*Isc) and equivalent impedance (Ze) = Esc/Isc, then equivalent reactance, Xe = square root of (Ze squared - Re squared). To get to the %Z on the nameplate, divide Ze by (nameplate voltage squared/nameplate VA).

Last, to your question, “So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?”

I’m note sure what predictions you are referring to here but I can speak to Utilites using lower impedance transformers to reduce losses. Yes, we do that for two reasons, First, it reduces losses. Losses which you as a customer pay for. We take the sum total of our generatEd energy and compare it to the sum total of our measured delivered energy and compute overall losses, the cost of which ultimately is paid for by the customer. So, reducing losses is a good thing, right? Second, reducing %Z raises the amount of short circuit current that a transformer may deliver. At the terminals of the transformer and assuming an infinite supply the available short circuit current = I (full load) / (%Z/100). So you can see the short circuit current gets higher as the equivalent impedance goes down.
 
Xptpcrewx is correct in that the open circuit test will only measure the losses due to hysteresis and eddy currents. Further, these losses only occur due to the current flow in the primary winding (primary in this sense means the source side of the transformer). The losses measured under the no-load condition are ignored for the most part to simplify calculations based on the very small loss of accuracy of the final result. It appears from your measurements you are attempting determine the %Z using the open circuit type of test. The percent Z represents a measurement of the equivalent resistance and reactance values of the transformer. That is it takes into account the primary and secondary resistance and reactance to give you an equivalent series impedance that you may use to determine the voltage regulation (voltage drop) that will occur based the load current and power factor.

An open circuit test will give you the no-load losses that are used with the short circuit test for the load losses, to calculate the overall efficiency of the transformer. I note here that the maximum efficiency of any transformer occurs when the open circuit test losses and the copper losses are equal, typically around 60% of the full load capacity.

To equate the %Z on the nameplate of the transformer to the actual equivalent %R and %X, you need to perform a short circuit test. The easiest path to this number is to determine the short circuit power (Psc), derive Re (equivalent Resistance) = Psc/(Isc*Isc) and equivalent impedance (Ze) = Esc/Isc, then equivalent reactance, Xe = square root of (Ze squared - Re squared). To get to the %Z on the nameplate, divide Ze by (nameplate voltage squared/nameplate VA).

Last, to your question, “So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?”

I’m note sure what predictions you are referring to here but I can speak to Utilites using lower impedance transformers to reduce losses. Yes, we do that for two reasons, First, it reduces losses. Losses which you as a customer pay for. We take the sum total of our generatEd energy and compare it to the sum total of our measured delivered energy and compute overall losses, the cost of which ultimately is paid for by the customer. So, reducing losses is a good thing, right? Second, reducing %Z raises the amount of short circuit current that a transformer may deliver. At the terminals of the transformer and assuming an infinite supply the available short circuit current = I (full load) / (%Z/100). So you can see the short circuit current gets higher as the equivalent impedance goes down.

Let me just emphasize that my wording may not have been clear or concise. When questioning/asking what can be inferred about losses by looking at the %Z, I mean looking at otherwise identical transformers and using their %Z as a comparative measure.

Here is good article my Eaton that touches on these issues:


Basically they state that due to the way efficiency is measured in the DOE standards, no load losses is the primary characteristic that the manufacturer wants to lower. Then they list the construction aspects that effect the core losses. The most common are:

1. Use of high-grade steel.
2. Lowering of the induction level on the core.
3. Use of different core constructions.

Then they state "Impedance changes" and state:

Although the impedance is not properly a characteristic of the core, it is closely related to the induction levels

So it seems there can be a relationship to %Z and no load losses, but I agree there is probably too many other factors to make an accurate generalization.
 
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