That percentage of primary or secondary rated voltage applied with other side shorted will cause full load current flowing on the other side.
Transformer Impedance?
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mbrooke
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This always equal to "classical" Z?
I agree.
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It is a trade off between material cost (higher for lower %Z) and operating cost (higher for higher %Z). At some point the customers are not willing to pay more for greater efficiency (lower %Z) just for the corresponding operating cost savings.
In states with strong energy codes, there may be specifications of max impedance in new transformer installations.
Julius Right
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Usually the power transformer construction way it is dictated by economical criteria. Mainly the iron to copper price ratio.
For a certain supply voltage V=K*Fe column Area*no of turns of copper conductor . From here if we try to reduce Fe column Area and so reducing iron price we have to increase no of turns*copper conductor and the copper price rise. There is a point somewhere where total price get a minimum.
That depends, of course, on what will be the price of iron against the copper at this time.
The leakage magnetic flux -the reactance- depends on square of no.of turns and it has to be limited due short-circuit current limitation in order to reduce the price of the equipment and conductors. ANSI Standard C37.010 and IEC 60076-5 limit this reactance according to transformer MVA.[Somewhere ±7.5% is allowed].An increased reactance will limit the current but will increase the voltage drop in the same time. On the other hand the total losses-iron and copper-has to be limited in order to reduce the energy delivery expenses.
Sometime there are construction limitation required. Then the leakage reactance could be affected.
Let's say you need to reduce the transformer height by 10% then the conductor bobbin thickness will be enlarged by 10%.In this case the leakage magnetic flux -the reactance- will rise some 14%. Φleak=k*N^2*(dcoil/3+s)/Lc where dcoil it is copper bobbin thickness s=distance in ambient N number of turns and Lc it is the bobbin length.
Lcnew=Lc*0.9 then dcoilnew=dcoilold/0.9.If dcoil≈s then:
Φleaknew=Φleak*(dcoilold/0.9/3+s)/dcoilold/3+s)/0.9=1.1389[13.89% more]
For a certain supply voltage V=K*Fe column Area*no of turns of copper conductor . From here if we try to reduce Fe column Area and so reducing iron price we have to increase no of turns*copper conductor and the copper price rise. There is a point somewhere where total price get a minimum.
That depends, of course, on what will be the price of iron against the copper at this time.
The leakage magnetic flux -the reactance- depends on square of no.of turns and it has to be limited due short-circuit current limitation in order to reduce the price of the equipment and conductors. ANSI Standard C37.010 and IEC 60076-5 limit this reactance according to transformer MVA.[Somewhere ±7.5% is allowed].An increased reactance will limit the current but will increase the voltage drop in the same time. On the other hand the total losses-iron and copper-has to be limited in order to reduce the energy delivery expenses.
Sometime there are construction limitation required. Then the leakage reactance could be affected.
Let's say you need to reduce the transformer height by 10% then the conductor bobbin thickness will be enlarged by 10%.In this case the leakage magnetic flux -the reactance- will rise some 14%. Φleak=k*N^2*(dcoil/3+s)/Lc where dcoil it is copper bobbin thickness s=distance in ambient N number of turns and Lc it is the bobbin length.
Lcnew=Lc*0.9 then dcoilnew=dcoilold/0.9.If dcoil≈s then:
Φleaknew=Φleak*(dcoilold/0.9/3+s)/dcoilold/3+s)/0.9=1.1389[13.89% more]
When multiplied by base z/100.
Efficiency is better and easier to use term to specify. Also variation in impedance may cause only slight corresponding variation in efficiency.
mbrooke
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I'm confused. As I understand power factor plays a major role, and a leading power factor can even raise Vs.
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Just a side note to a very informative thread:
IEEE 519, the “harmonics” spec, is generally regarded as applying to the end user in terms of requiring us to take measures to reduce the THD we create. But IEE 519 also applies to the utility side as well, dictating that the utility must maintain a minimum amount of system impedance in order to help mitigate the propagation of harmonic distortion between users. Without that, utilities could reduce their system losses by lowering impedance, but the effect would harm the end users from a harmonic distortion perspective, increasing the measures they must take.
IEEE 519, the “harmonics” spec, is generally regarded as applying to the end user in terms of requiring us to take measures to reduce the THD we create. But IEE 519 also applies to the utility side as well, dictating that the utility must maintain a minimum amount of system impedance in order to help mitigate the propagation of harmonic distortion between users. Without that, utilities could reduce their system losses by lowering impedance, but the effect would harm the end users from a harmonic distortion perspective, increasing the measures they must take.
Impedance is dropping
Impedance is dropping
High impedance transformers have less available short circuit current. But low impedance transformers are in general more efficient. My experience is from calculating short circuit current - so not relative to constructing the transformer.
The books that were published a few years ago with typical impedance are no longer correct. I was using 5% when I didn't know for an oil filled utility transformer and got specs on a local transformer and it was 1.6%. My calculations were using an infinite bus and no series rating - so likely were okay, but the change upset me quite a bit. I had gotten utility transformer information previously and compared it with my book and it had been appropriate. No more.
Our local utility gave me a list of all the transformers and impedance that they were using as of two years ago. For 500KVA 208V output, I need to use 1.6% to be safe. For the larger transformers (over 500KVA), I need to use 3%. The utility generally does not want to commit on typical jobs until after the transformer has been tagged for delivery. That generally is too late for me to be specifying the distribution panels.
I realize that you are in a much larger range - but my thought is that the higher impedance makes the other equipment less expensive due to lower fault ratings.
Impedance is dropping
High impedance transformers have less available short circuit current. But low impedance transformers are in general more efficient. My experience is from calculating short circuit current - so not relative to constructing the transformer.
The books that were published a few years ago with typical impedance are no longer correct. I was using 5% when I didn't know for an oil filled utility transformer and got specs on a local transformer and it was 1.6%. My calculations were using an infinite bus and no series rating - so likely were okay, but the change upset me quite a bit. I had gotten utility transformer information previously and compared it with my book and it had been appropriate. No more.
Our local utility gave me a list of all the transformers and impedance that they were using as of two years ago. For 500KVA 208V output, I need to use 1.6% to be safe. For the larger transformers (over 500KVA), I need to use 3%. The utility generally does not want to commit on typical jobs until after the transformer has been tagged for delivery. That generally is too late for me to be specifying the distribution panels.
I realize that you are in a much larger range - but my thought is that the higher impedance makes the other equipment less expensive due to lower fault ratings.
tbuller
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Transformer Impedance affects fault current
Transformer Impedance affects fault current
Transformer impedance affects fault current. It impacts the rating of Switchgear, MCC's, panels etc. It also impacts arc flash levels and motor starting. MCC's and Switchgear have mechanical bracing to keep them from breaking apart during a fault.
When buying a new transformer it is important to calculate fault current to make sure your MCC's and Switchgear can handle it. For a transformer with a secondary Full Load Amp (FLA) rating of 1,200 Amps and 5% impedance the calculation would look like the following: FLA x 100 / %impedance.
1,200 x 100 / 5 = 24,000A.
A lot of panels and MCC's are only rated for 8,000 - 12,000 Amps so it is a good rule of thumb to make sure the equipment that you specify has an Asymmetrical Fault current rating that is a least 20% higher than what you calculated.
Transformer Impedance affects fault current
Transformer impedance affects fault current. It impacts the rating of Switchgear, MCC's, panels etc. It also impacts arc flash levels and motor starting. MCC's and Switchgear have mechanical bracing to keep them from breaking apart during a fault.
When buying a new transformer it is important to calculate fault current to make sure your MCC's and Switchgear can handle it. For a transformer with a secondary Full Load Amp (FLA) rating of 1,200 Amps and 5% impedance the calculation would look like the following: FLA x 100 / %impedance.
1,200 x 100 / 5 = 24,000A.
A lot of panels and MCC's are only rated for 8,000 - 12,000 Amps so it is a good rule of thumb to make sure the equipment that you specify has an Asymmetrical Fault current rating that is a least 20% higher than what you calculated.
You either have to live with rule of thumbs or compute for the asymmetry of your fault current. How much your initial fault current will be depends on the X/R of the system, and not just a simple rule of thumb!
mbrooke
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This is what "raises" the since wave above the zero crossing? Or just the DC component? Or both?
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