Calculations Involving Transformer Impedance

[jlaturell]

I'm trying to get back to basics. Over the years, Mike has taught numerous electricians, including my late father, that if you want to find the value of amperage flowing in a circuit, you've got know what electromotive force and what opposition to that force is also involved. While studying Ohms Law and AC circuitry long ago , I learned that impedance is the vector sum of resistance plus reactance, whose value is displayed in ohms. Transformers obviously contain wire coils. AC applied to any wire coil type will induce some amount of inductive reactance (X_L) into the circuit in addition to the wire resistance (R). Today, transformer nameplate data displays transformer impedance as a "percentage". Percentage of what??? How is this percentage value converted back into basic Z??? Why don't AIC calculations simply use I (amps), E (volts), and in this case Z (ohms)??? Thank you for any reply or consideration.

P.S. Yes, this thread will probably also evolve into later inquries/discussions about pu values.

 

[iceworm]

... Today, transformer nameplate data displays transformer impedance as a "percentage". Percentage of what??? How is this percentage value converted back into basic Z??? Why don't AIC calculations simply use I (amps), E (volts), and in this case Z (ohms)??? ...

P.S. Yes, this thread will probably also evolve into later inquries/discussions about pu values.

Good morning jl -

I have attached two pages from the IEEE Violet book (IEEE 551) that give a pretty good, short, explanation on per unit and why.

Perhaps this will get you started.

ice

 

[jim dungar]

Today, transformer nameplate data displays transformer impedance as a "percentage". Percentage of what??? How is this percentage value converted back into basic Z??? Why don't AIC calculations simply use I (amps), E (volts), and in this case Z (ohms)??? Thank you for any reply or consideration.

%Z is more correctly called %IZ. Following Ohm's law, V=IZ, %Z is really a %V kind of confusing but please hold on.

First. you take the scecondary of a transformer and wire it with a short ciruit. This effectively results in the only impedance bing that of the transformer.
Second measure the current that flows through the short circuited wires this gives us an I.
Third slowly increase the voltage into the primary of the transformer until you have 100% of the rated current flowing through the short circuit. You now have a voltage measurement.

We now have a percentage of rated voltage that creates 100 percent of rated current flowing through a fixed impedance. Using Ohm's laws, %V=%I%Z. Because we are at 100% of current %I = 1 and our equation becomes %V=%Z.

So %Z is actually the amount of voltage that will cause 1 rated unit of current to flow through the impedance of the secondary.

 

[mivey]

%Z is more correctly called %IZ. Following Ohm's law, V=IZ, %Z is really a %V kind of confusing but please hold on.

First. you take the scecondary of a transformer and wire it with a short ciruit. This effectively results in the only impedance bing that of the transformer.
Second measure the current that flows through the short circuited wires this gives us an I.
Third slowly increase the voltage into the primary of the transformer until you have 100% of the rated current flowing through the short circuit. You now have a voltage measurement.

We now have a percentage of rated voltage that creates 100 percent of rated current flowing through a fixed impedance. Using Ohm's laws, %V=%I%Z. Because we are at 100% of current %I = 1 and our equation becomes %V=%Z.

So %Z is actually the amount of voltage that will cause 1 rated unit of current to flow through the impedance of the secondary.

The test circuit used to get %Z is correct. It is a version of the Thevenin circuit with specific test values. That does not mean we rename the impedance to something else.

%Z is NOT more correctly called %IZ. It is a measure of the impedance and is most correctly called %Z. Even though the %I = 1 it is still %Z = %V / %I = %V / 1

In other words, you could call %Z by the relationship %V/%I and that would be more correct than saying %Z is more correctly called %I%Z.

 

[mivey]

25 kVA example

25 kVA example

Given:
25 kVA = Sbase
7200 V_primary = Vbase_pri
240 V_secondary = Vbase_sec
%Z = 3%

We have:
Ibase_sec = Sbase / Vbase = 25 kVA / 0.24 kV = 104.17 A
Ibase_pri = Sbase / Vbase = 25 kVA / 7.2 kV = 3.472 A
Zbase_sec = Vbase_sec / Ibase_sec = 2.304 Ω
Zbase_p = Vbase_pri / Ibase_pri = 2073.6 Ω
Zsec = %Z * Zbase = 0.03 * 2.304 Ω = 0.06912 Ω
Zpri = %Z * Zbase = 0.03 * 2073.6 Ω = 62.208 Ω



As for test inputs:

Voltage Input Test at Rated Current:

Irated_sec = 104.17 A = 100.00% Ibase_sec
Vsec = 7.2 V = 3.00% Vbase_sec
Vpri = 216 V = 3.00% Vbase_pri
Ipri = 3.472 A = 100.00% Ibase_pri
Power = 750W => practical test
Zcalc_sec = V/I = 7.2 V / 104.17 A = 0.06912 Ω => 0.06912 Ω / 2.304 Ω = 3.00% Zbase_sec = %V/%I
Zcalc_pri = V/I = 216 V / 3.472 A = 62.208 Ω => 62.208 Ω / 2073.6 Ω = 3.00% Zbase_pri = %V/%I

note that for this case, %Z = %V/%I = %V/1



Current Input Test at Rated Voltage:

Vrated_sec = 240 V = 100.00% Vbase_sec
Isec = 3472.22 A = 3333.33% Ibase_sec (i.e., we need full fault current!)
Ipri = 115.74 A = 3333.33% Ibase_pri
Vpri = 7200 V = 100.00% Vbase_pri
Power = 833,333 W => test at fault current is not a practical test
Zcalc_sec = V/I = 240 V / 3472.22 A = 0.06912 Ω => 0.06912 Ω / 2.304 Ω = 3.00% Zbase_sec = %V/%I
Zcalc_pri = V/I = 7200 V / 115.74 A = 62.208 Ω => 62.208 Ω / 2073.6 Ω = 3.00% Zbase_pri = %V/%I

note that for this case, %Z = %V/%I = 1/%I


Current Input Test at < Rated Voltage:

Vinput_sec = 2.4 V = 1.00% Vbase_sec
Isec = 34.72 A = 33.33% Ibase_sec
Ipri = 1.1574 A = 33.33% Ibase_pri
Vpri = 72 V = 1.00% Vbase_pri
Power = 83.33 W => could be a practical test, but not the way it is normally done
Zcalc_sec = V/I = 2.4 V / 34.7222 A = 0.06912 Ω => 0.06912 Ω / 2.304 Ω = 3.00% Zbase_sec = %V/%I
Zcalc_pri = V/I = 72 V / 1.1574 A = 62.208 Ω => 62.208 Ω / 2073.6 Ω = 3.00% Zbase_pri = %V/%I

note that for this case, %Z = %V/%I



You can see that the measurement of V/I gives us Z and %V/%I = %Z. Setting the test %I to 100% does not mean that %Z is more correctly called %IZ any more than setting the test %V to 100% would mean %Z is more correctly called %V/Z.

%Z is most correctly called %Z.

 

[jlaturell]

Gentemen;
Thank you both so very much for your explanations. Mivey, your example gives me something to scrutinize/"chew on" and is helping to clear out the mud in my mind's eye somewhat. I promise to follow your examples someday by helping others who simply want to learn and understand.

I've been servicing generators/switchboards for a few years. This curiosity started by trying to understand how circuit breaker L,S,I, and G setpoints are derived from SKM coordination studies. I still have several questions, but your replies herein will answer many of them.

I sometimes wonder why information like this appears so difficult to obtain. Perhaps it is merely one more product of the post Industrial Revolution. More than likely however, it's just taken for granted and seems to be a sad disinterest by so many Americans in the power industry to truly understand what all these available software apps are doing. Keep up your great work!

Teach it once --- Learn it twice.

 

[kingpb]

You can get the individual R and X components by knowing Z and the transformer X/R ratio. It can be calculated in % or ohms using the following relationship:

Z = (R² + X²)^1/2

 

[mivey]

More than likely however, it's just taken for granted and seems to be a sad disinterest by so many Americans in the power industry to truly understand what all these available software apps are doing.

I see that more and more. Many depend on the results of some software with no understanding of how to verify what the computer tells them. They look like a deer in headlights when something unexpected happens. That is usually when the excuses start to fly.

 

[don_resqcapt19]

... Many depend on the results of some software with no understanding of how to verify what the computer tells them. ...

Even worse is when they have no idea that the answer they got from the software program not a realistic answer.

 

[iceworm]

You can get the individual R and X components by knowing Z and the transformer X/R ratio. ...

Q: How does the mfg measure the X/R ratio?

If you are going to tell me they measure the X and the R and divide the two:slaphead: Then I will ask, why don't they give the X and R values?

I've never given it much thought other than to verify CB ratings are within spec to interupt a transformer SCC. As I recall, most CBs are rated for an X/R < 10.

ice

 

[jlaturell]

I see that more and more. Many depend on the results of some software with no understanding of how to verify what the computer tells them. They look like a deer in headlights when something unexpected happens. That is usually when the excuses start to fly.

Amen to that; And when you're the low man on the totem pole as I was recently, the flying excuses turn into finger-pointing which resulted in getting thrown under the bus -- not fun.

 

[jim dungar]

%Z is most correctly called %Z.

Take it up with IEEE and other industry publications for their use of "impedance voltage", for example:

The percent impedance voltage levels considered as standard for two-winding transformers are listed in tables 10-15 and 10-16, and a value specified above or below those listed may result in higher costs. The percent impedance voltage of a two-winding transformer shall have a tolerance of 7.5% of the specified value.

 

[jlaturell]

Does one need to be independently wealthy, join IEEE, and purchase these pubs, or is there a secret handshake that someone like me can learn so I might download them for personal use only and learn from them?

 

[jim dungar]

Does one need to be independently wealthy, join IEEE, and purchase these pubs, or is there a secret handshake that someone like me can learn so I might download them for personal use only and learn from them?

I found an employer that was willing to get them for me. For students, the engineering libraries usually can access them.

 

[mivey]

Then I will ask, why don't they give the X and R values?

They do, in the test report. I've always wished they would have put them on the nameplate because finding the original test report is not always easy.

 

[mivey]

Take it up with IEEE and other industry publications for their use of "impedance voltage", for example:

That's all fine and good, but we still seek the numerically equivalent %Z.

From Westinghouse's Electric Utility Engineering Reference Book Volume 3: "Distribution Systems", 1959:

The impedance usually given for a transformer is the equivalent series impedance...The percent impedance of a transformer can be found by short circuiting one winding and applying sufficient voltage to the other winding to circulate rated current through the transformer. The voltage required to circulate rated current, expressed as a percentage of the rated voltage of the winding to which the voltage is applied, is numerically equal to the percent impedance of the transformer.

So while we can discuss the numerically equivalent %V, or %IZ, or even the %kVA, we are testing to find the percent impedance, %Z.

 

[kingpb]

Take it up with IEEE and other industry publications for their use of "impedance voltage", for example:

IEEE 100 defines Impedance Voltage as:
The voltage required to circulate rated current through one of two specified windings of a transformer when the other winding is short-circuited,
with the windings connected as for rated voltage operation.

Note: It is usually expressed in per unit, or percent, of the rated voltage of the winding in which the voltage is measured.

 

[Hv&Lv]

Given:
25 kVA = Sbase
7200 V_primary = Vbase_pri
240 V_secondary = Vbase_sec
%Z = 3%

We have:
Ibase_sec = Sbase / Vbase = 25 kVA / 0.24 kV = 104.17 A
Ibase_pri = Sbase / Vbase = 25 kVA / 7.2 kV = 3.472 A
Zbase_sec = Vbase_sec / Ibase_sec = 2.304 Ω
Zbase_p = Vbase_pri / Ibase_pri = 2073.6 Ω
Zsec = %Z * Zbase = 0.03 * 2.304 Ω = 0.06912 Ω
Zpri = %Z * Zbase = 0.03 * 2073.6 Ω = 62.208 Ω



As for test inputs:

Voltage Input Test at Rated Current:

Irated_sec = 104.17 A = 100.00% Ibase_sec
Vsec = 7.2 V = 3.00% Vbase_sec
Vpri = 216 V = 3.00% Vbase_pri
Ipri = 3.472 A = 100.00% Ibase_pri
Power = 750W => practical test
Zcalc_sec = V/I = 7.2 V / 104.17 A = 0.06912 Ω => 0.06912 Ω / 2.304 Ω = 3.00% Zbase_sec = %V/%I
Zcalc_pri = V/I = 216 V / 3.472 A = 62.208 Ω => 62.208 Ω / 2073.6 Ω = 3.00% Zbase_pri = %V/%I

note that for this case, %Z = %V/%I = %V/1



Current Input Test at Rated Voltage:

Vrated_sec = 240 V = 100.00% Vbase_sec
Isec = 3472.22 A = 3333.33% Ibase_sec (i.e., we need full fault current!)
Ipri = 115.74 A = 3333.33% Ibase_pri
Vpri = 7200 V = 100.00% Vbase_pri
Power = 833,333 W => test at fault current is not a practical test
Zcalc_sec = V/I = 240 V / 3472.22 A = 0.06912 Ω => 0.06912 Ω / 2.304 Ω = 3.00% Zbase_sec = %V/%I
Zcalc_pri = V/I = 7200 V / 115.74 A = 62.208 Ω => 62.208 Ω / 2073.6 Ω = 3.00% Zbase_pri = %V/%I

note that for this case, %Z = %V/%I = 1/%I


Current Input Test at < Rated Voltage:

Vinput_sec = 2.4 V = 1.00% Vbase_sec
Isec = 34.72 A = 33.33% Ibase_sec
Ipri = 1.1574 A = 33.33% Ibase_pri
Vpri = 72 V = 1.00% Vbase_pri
Power = 83.33 W => could be a practical test, but not the way it is normally done
Zcalc_sec = V/I = 2.4 V / 34.7222 A = 0.06912 Ω => 0.06912 Ω / 2.304 Ω = 3.00% Zbase_sec = %V/%I
Zcalc_pri = V/I = 72 V / 1.1574 A = 62.208 Ω => 62.208 Ω / 2073.6 Ω = 3.00% Zbase_pri = %V/%I

note that for this case, %Z = %V/%I



You can see that the measurement of V/I gives us Z and %V/%I = %Z. Setting the test %I to 100% does not mean that %Z is more correctly called %IZ any more than setting the test %V to 100% would mean %Z is more correctly called %V/Z.

%Z is most correctly called %Z.

I was reading some more on the O.C. Seever book, and was wondering about this today. Actually thought about PM'ing you to ask this question. Glad it came up today.
I see that you took 100%/3%, and came up with 33.33 times the nameplate rating for the fault current calculation. (33.33X104.17)

And from what I am reading here, if I had an XF with %Z=100, then 100% / 100% =1, so one times the nameplate rating will be the fault amps, or 104.17 fault amps at 100% Z.
I realize 100% impedance is silly, but I am trying to get a good handle on this. I was told by someone that 100% impedance equals no current. According to the math, I disagree. Of course, I've been wrong before.

 

[mivey]

And from what I am reading here, if I had an XF with %Z=100, then 100% / 100% =1, so one times the nameplate rating will be the fault amps, or 104.17 fault amps at 100% Z.

Correct

I realize 100% impedance is silly, but I am trying to get a good handle on this. I was told by someone that 100% impedance equals no current. According to the math, I disagree. Of course, I've been wrong before.

It certainly does not equal zero current. What it does mean is that at full current, there will be zero voltage because all of the voltage is dropped across the internal series impedance.

Consider the 25 kVA example with Zbase_sec = 2.304 Ω. With shorted terminals (zero Ω load) we get:
I = Vs / (Zt + Zload) = 240 V / (2.304 Ω + 0 Ω) = 104.17 A
Voutput = Vs - I * Zt = 240 V - 104.17 A * 2.304 Ω = 0 V

But it does not mean zero volts for all cases.

Consider the 25 kVA example with Zbase_sec = 2.304 Ω. With a 2 Ω load we get:
I = Vs / (Zt + Zload) = 240 V / (2.304 Ω + 2 Ω) = 55.762 A
Voutput = Vs - I * Zt = 240 V - 55.762 A * 2.304 Ω = 240 V - 128.476 V = 111.52 V

 

[jlaturell]

I was reading some more on the O.C. Seever book, and was wondering about this today.

Are you refering to this book???

Power Systems Handbook: Design, Operation & Maintenance,

ISBN	0136824935, 9780136824930

Thank you for any reply.




Teach it once -- Learn it twice.