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Transformer without secondary 1
- Thread starter edison123
- Start date
Hi edison,
Interesting question. As far as I can see, it will just behave as a reactor connected across the supply. There will be no mutual flux coupling because there is only one winding with two terminals, so the standard reactance equations for an iron cored inductor will apply. The iron loss will stay the same regardless of the presence or not of the secondary winding.
Out of sheer curiousity, is this just idle thoughts or a real situation?
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I don't suffer from insanity. I enjoy it...
Interesting question. As far as I can see, it will just behave as a reactor connected across the supply. There will be no mutual flux coupling because there is only one winding with two terminals, so the standard reactance equations for an iron cored inductor will apply. The iron loss will stay the same regardless of the presence or not of the secondary winding.
Out of sheer curiousity, is this just idle thoughts or a real situation?
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- Moderator
- #3
Hello ScottyUK and edison123
I agree with you scottyUK, with three exceptions.
A single phase transforme with the secondaries in parallel should have no circulating current. If however, the secondary windings are not perfectly balanced there will be a small circulating current. This will add a small loss that will be removed if the winding is removed.
If the secondary was delta connected and was carrying any harmonic currents, there will be a slight reduction in losses if the secondary is removed.
If the primary is 4-wire wye connected and the secondary is delta connected there are excessive currents when the primary voltage is unbalanced. The circulating currents and the resulting losses can exceed the full load losses. (That's probably why the RUS standards warn that this connection can result in transformer burnout.)
I understand that some three-phase transformers use a core configuration that results in harmonic fluxes that need a delta winding to provide a return path. Hence tertiary windings.
I'm open to comments and correction on this point.
For most transformers you are correct scottyUK. I believe that for all single-phase transformers a series connected secondary has no effect on the no load losses.
For three phase Wye Delta transformers, depending on the core arrangement, harmonic currents and the primary voltage balance, the removal of the delta winding, (Whether secondary or tertiary), may result in the same losses, less losses, a lot less losses, or more losses.
The definitive word, to quote one of our other contributors, is "It Depends".
respectfully
I agree with you scottyUK, with three exceptions.
A single phase transforme with the secondaries in parallel should have no circulating current. If however, the secondary windings are not perfectly balanced there will be a small circulating current. This will add a small loss that will be removed if the winding is removed.
If the secondary was delta connected and was carrying any harmonic currents, there will be a slight reduction in losses if the secondary is removed.
If the primary is 4-wire wye connected and the secondary is delta connected there are excessive currents when the primary voltage is unbalanced. The circulating currents and the resulting losses can exceed the full load losses. (That's probably why the RUS standards warn that this connection can result in transformer burnout.)
I understand that some three-phase transformers use a core configuration that results in harmonic fluxes that need a delta winding to provide a return path. Hence tertiary windings.
I'm open to comments and correction on this point.
For most transformers you are correct scottyUK. I believe that for all single-phase transformers a series connected secondary has no effect on the no load losses.
For three phase Wye Delta transformers, depending on the core arrangement, harmonic currents and the primary voltage balance, the removal of the delta winding, (Whether secondary or tertiary), may result in the same losses, less losses, a lot less losses, or more losses.
The definitive word, to quote one of our other contributors, is "It Depends".
respectfully
I wouldn't expect much difference, but the secondary winding would have a voltage developed on it and there will be some slight leakage current even if it is open-circuited. The slight energy required to maintain the electric field in the secondary windings has to come from somewhere and that would be from the primary.
So I would actually expect the primary current to be (very) slightly less if the secondary winding was removed. But then again, I never did too well in electro-magnetics.
So I would actually expect the primary current to be (very) slightly less if the secondary winding was removed. But then again, I never did too well in electro-magnetics.
- Thread starter
- #5
Hi scotty & waross,
I am making in-house a single phase, HV test trafo. I wanted to check my design calculation of no-load current before finalizing the no. of turns.
Since trafo is always cited as induction motor without air-gap (or vice versa, if you wish) and since induction motor no-load current increases drastically without the rotor (due to lack of back emf), I was wondering would that apply to trafos also.
I will be doing this experiment (???) next week and will be back here with the results.
* Basically, I would like a full-time job on part-time basis *
I am making in-house a single phase, HV test trafo. I wanted to check my design calculation of no-load current before finalizing the no. of turns.
Since trafo is always cited as induction motor without air-gap (or vice versa, if you wish) and since induction motor no-load current increases drastically without the rotor (due to lack of back emf), I was wondering would that apply to trafos also.
I will be doing this experiment (???) next week and will be back here with the results.
* Basically, I would like a full-time job on part-time basis *
- Moderator
- #6
Hello all;
Actually dpcs post may be the most valid. If the primary winding is the low voltage and the secondary voltage is extremely high, you may have some resistive and capacitive losses on the secondary. I don't think the losses will be significant, but what are your design voltages and VA ratings.
The smaller your transformer is, the higher the secondary no-load percentage loss may be.
The formula for no load losses depends on the current, and on a distribution transformer this is taken at zero at no load.
If you are considering a secondary of several thousand volts, you may want to do a spreadsheet model of the capacitance of part of the high voltage winding.
First; I would suggest calculating the capacity of one turn of the high voltage winding on the bottom layer. Enter this value in a cell of an excel spreadsheet. Then calculate the capacitive current based on the average voltage to ground of this turn. (For a high voltage transformer I would ignore the secondary voltage and consider the secondary current to be at ground potential. The small error will be on the conservative side.)
Now increment or decrement the voltage value by the turn-to-turn voltage in the lower cells. Sum the results.
Second; Calculate the capacity of a turn in the first layer to an adjacent turn in the second layer. Calculate the capacitive current based the turn-to-turn voltage.
In the following cells increment the voltage by twice the turn-to-turn voltage. Sum the results.
Third; Repeat the second step for succeeding layers.
Fourth; Calculate the turn-to-turn capacitance. Calculate the current based on the turn-to-turn voltage. Multiply by the number of turns.
Fifth; Sum the sums.
This is a rough approximation of the capacitive current.
Errors; This is most valid for perfectly laid turns. With less than perfect alignment of the turns the current will be less than calculated. Your calculation should be the worst case. The actual current will be less than the calculated current.
The capacitance you calculate for the first layer will be subject to assumptions and will not be precise. Don't worry.
The turn-to-turn and layer-to-layer capacitances will be more precise but still not particularly accurate. Don't worry. Even if you are off by 300% in either direction. It may not matter. Consider; if your calculations indicate that the capacitive losses are 3% of the allowable losses, does it matter if the real figure is 1% of allowable losses or 9% of allowable losses?
If you are able to estimate the resistance of your insulation, from turn to turn, you can use a similar process to estimate the resistive losses.
This doesn't have much to do with flux. It is the effect of the voltage stress on the insulation (Capacitance) and the resistive current through the insulation.
Whether you do the calculations or not, we will be interested to know if there is a measurable difference in the no load current before and after you install the secondary winding.
BTW meggering the windings will be interesting.
respectfully
Actually dpcs post may be the most valid. If the primary winding is the low voltage and the secondary voltage is extremely high, you may have some resistive and capacitive losses on the secondary. I don't think the losses will be significant, but what are your design voltages and VA ratings.
The smaller your transformer is, the higher the secondary no-load percentage loss may be.
The formula for no load losses depends on the current, and on a distribution transformer this is taken at zero at no load.
If you are considering a secondary of several thousand volts, you may want to do a spreadsheet model of the capacitance of part of the high voltage winding.
First; I would suggest calculating the capacity of one turn of the high voltage winding on the bottom layer. Enter this value in a cell of an excel spreadsheet. Then calculate the capacitive current based on the average voltage to ground of this turn. (For a high voltage transformer I would ignore the secondary voltage and consider the secondary current to be at ground potential. The small error will be on the conservative side.)
Now increment or decrement the voltage value by the turn-to-turn voltage in the lower cells. Sum the results.
Second; Calculate the capacity of a turn in the first layer to an adjacent turn in the second layer. Calculate the capacitive current based the turn-to-turn voltage.
In the following cells increment the voltage by twice the turn-to-turn voltage. Sum the results.
Third; Repeat the second step for succeeding layers.
Fourth; Calculate the turn-to-turn capacitance. Calculate the current based on the turn-to-turn voltage. Multiply by the number of turns.
Fifth; Sum the sums.
This is a rough approximation of the capacitive current.
Errors; This is most valid for perfectly laid turns. With less than perfect alignment of the turns the current will be less than calculated. Your calculation should be the worst case. The actual current will be less than the calculated current.
The capacitance you calculate for the first layer will be subject to assumptions and will not be precise. Don't worry.
The turn-to-turn and layer-to-layer capacitances will be more precise but still not particularly accurate. Don't worry. Even if you are off by 300% in either direction. It may not matter. Consider; if your calculations indicate that the capacitive losses are 3% of the allowable losses, does it matter if the real figure is 1% of allowable losses or 9% of allowable losses?
If you are able to estimate the resistance of your insulation, from turn to turn, you can use a similar process to estimate the resistive losses.
This doesn't have much to do with flux. It is the effect of the voltage stress on the insulation (Capacitance) and the resistive current through the insulation.
Whether you do the calculations or not, we will be interested to know if there is a measurable difference in the no load current before and after you install the secondary winding.
BTW meggering the windings will be interesting.
respectfully
- Thread starter
- #7
Here are the test results
No-load primary current without the secondary winding - 2.0 A and with the secondary winding - 1.1 A.
To some extent, it does resemble induction motor, I guess.
* Basically, I would like a full-time job on part-time basis *
No-load primary current without the secondary winding - 2.0 A and with the secondary winding - 1.1 A.
To some extent, it does resemble induction motor, I guess.
* Basically, I would like a full-time job on part-time basis *
- Moderator
- #8
That is strange. I understood that the current increase in a motor with the rotor removed was due to the extremely large air gap where the rotor used to be.
One explanation may be mechanical. If for some reason there was a slight air that was closed up or reduced when the secondary winding was installed that would explain the drop in current.
The increasing magnetic field in the core induces the back EMF in the primary winding. Any change in the secondary circuit should increase the current, not reduce it.
I hope someone can explain this to us.
respectfully.
One explanation may be mechanical. If for some reason there was a slight air that was closed up or reduced when the secondary winding was installed that would explain the drop in current.
The increasing magnetic field in the core induces the back EMF in the primary winding. Any change in the secondary circuit should increase the current, not reduce it.
I hope someone can explain this to us.
respectfully.
I wonder if you have introduced a small airgap into the core structure during reassembly? Did you have to remove a limb of the core to extract the winding, or did you do it the hard way? If the core didn't go back together perfectly it might explain the result becasue the magnetic properties would change - I can't see why removal of the secondary alone ould cause a change in mag current.
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I don't suffer from insanity. I enjoy it...
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- Thread starter
- #10
- Moderator
- #11
From your currents I would suspect that in the first test the top yoke was slightly out of place or had a bit of foriegn matter under it.
Scotty, we are in agreement about the probable cause but not when. Given that the first test without the primary winding had more current, I think that there may have been a small accidental air gap during the first test and it's good now.
Did I misunderstand your post, or am I confused.
Scotty, we are in agreement about the probable cause but not when. Given that the first test without the primary winding had more current, I think that there may have been a small accidental air gap during the first test and it's good now.
Did I misunderstand your post, or am I confused.
Hi waross,
Yes, we're agreeing: the test resulting in the smaller current probably had a better magnetic circuit with a smaller unintentional air gap being present.
My guess was that prior to the transformer being dismantled it probably had a good magnetic circuit. When the secondary was removed and the core re-assembled I suspect that the core did not go back together quite as well as it did at original manufacture. The change could be due to reduced clamping force or surface roughness of the interlamination insulation varnish.
What is the transformer rating?
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I don't suffer from insanity. I enjoy it...
Yes, we're agreeing: the test resulting in the smaller current probably had a better magnetic circuit with a smaller unintentional air gap being present.
My guess was that prior to the transformer being dismantled it probably had a good magnetic circuit. When the secondary was removed and the core re-assembled I suspect that the core did not go back together quite as well as it did at original manufacture. The change could be due to reduced clamping force or surface roughness of the interlamination insulation varnish.
What is the transformer rating?
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1
- #13
CarlPugh
Electrical
- Jul 16, 2002
- 142
?Do you have a very high voltage secondary?
waross mentioned this possibility in his posting.
High voltage secondaries have a capacitive current that cancels the core inductive current, so that the total current is less.
The book
Electronic Transformers and Circuits
Reuben Lee, Leo Wilson, Charles E. Carter
gives a method for calculating capacitive current.
A good indication of capacitive current is if when the input voltage is slowly raised, the exciting current starts increasing normally and then drops as the voltage increases and then again starts increasing at a higher voltage.
waross mentioned this possibility in his posting.
High voltage secondaries have a capacitive current that cancels the core inductive current, so that the total current is less.
The book
Electronic Transformers and Circuits
Reuben Lee, Leo Wilson, Charles E. Carter
gives a method for calculating capacitive current.
A good indication of capacitive current is if when the input voltage is slowly raised, the exciting current starts increasing normally and then drops as the voltage increases and then again starts increasing at a higher voltage.
- Thread starter
- #14
This is a 400/30000 V HV test trafo.
With secondary winding (Open circuit)
Primary Volts Primary Current
250 0.2
300 0.2
350 0.4
400 1.1
Without secondary winding
Primary Volts Primary Current
175 0.10
200 0.12
225 0.14
250 0.15
275 0.18
300 0.28
325 0.50
350 0.89
375 1.40
400 2.00
* Basically, I would like a full-time job on part-time basis *
With secondary winding (Open circuit)
Primary Volts Primary Current
250 0.2
300 0.2
350 0.4
400 1.1
Without secondary winding
Primary Volts Primary Current
175 0.10
200 0.12
225 0.14
250 0.15
275 0.18
300 0.28
325 0.50
350 0.89
375 1.40
400 2.00
* Basically, I would like a full-time job on part-time basis *
- Moderator
- #15
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