Transformer-Rectifier output inductance 2

[jimbofitz]

A customer is asking about the inductance seen from the output terminals of a transformer rectifier system. It is a 24 pulse-rectifier comprised of 4 phase shifted parallel diode bridges. We use 3 interphase transformers to obtain 24 pulse operation. There also is an output inductor. I am having trouble with the concept of the IPT inductance during a short circuit. They pass DC current but if we try to interrupt a fault on the output the customer is worried about the engery stored (snubbers) and the fault current rate of rise. What do i need to consider to properly understand/predict for this? In other wordds, what would i see if there is a bolted short circuit on the DC bus for a A/us rise time? When the overcurrent protection trips what do i need to consider for energy stored?

Thank you in advance for your assistance, i hope you guys can point me in the right direction as these sytems are new to me.

Regards,
Jim

 

[Skogsgurra]

The basic answer is that your current will rise according to U/L amps/second. Make U equal to DC mean value, which is very close to the AC RMS times sqrt(2) in a 24 pulse rectifier. And L is inductance of the reactor.

The stored energy in the inductor is 0.5*L*I^2 and the time constant is (neglecting diode voltage drop) equal to L/R, where R is resistance of reactor (bolted short and short cables between diodes and reactor).

Using this simplified calculation, you will get the worst (highest) result. If your reactor is saturating at some high DC current level - which is very common - then your rate of rise will be higher after saturation point and the stored energy will also deviate from calculated value.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[jimbofitz]

Skogsgurra:

Perhaps it is my inexperience showing but what is U in this discussion? Are you referring to the DC values as per unit?

You make a good point about energy storage changing after saturation, I hadn't thought of that yet.

 

[Skogsgurra]

No, U is the DC voltage. It is very close to AC peak in a 24 pulse rectifier.

Of course, you can use p.u. if that is what you are used to. I usually use engineering units for everyday things. Per unit was back in school. And that's a long time ago.

There is one more thing to consider regarding rate of rise for current. The transformer impedance will reduce the rate of rise. But again, the simple result will always be safe.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[jimbofitz]

Oh ok I don't use per unit unless someone forces me. I haven't really heard the DC voltage referred to as U, is that common outside the US?

Are we talking about the inductance of the parallel combination of the interphase reactors + the output inductor, that is what is troubling me the most?

 

[Skogsgurra]

I think that a schematic would help. It is not quite clear how your set up is.

Are you using center tapped reactors?

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[jimbofitz]

Please refer to my orignal post, I included a schematic there.

Yes, they are center tapped.

 

[electricuwe]

The important message is that the interphase transformer is a transformer not an inductor. Therefore it does not store significant energy.

 

[jimbofitz]

From my understanding they aren't acting as a transformer in this capacity. In a balanced system they are transparent to the DC because the fluxes cancel. It is the AC voltage imbalances due to leakage reactance and integer-turn limitations of the phase shifting transformer that will be smoothed by the reactor. They are a gapped center-tapped reactor which will contain some energy when the overcurrent is interuppted in a short time. The customer has only indicated shutting down the DC side of the fault, nothing has been discussed about removing AC input power at this time. This means my circuit will be energized throughout the event.

My problem is that i can't figure out the impedance looking into the DC terminals with my current configuration.

 

[Skogsgurra]

Sorry. I missed that diagram.

Your intergroup reactors make my earlier numbers a little less. The voltage is somewhat lower than I said, but if you use the AC peak voltage, you will get a "safe error" i.e. you will get a somewhat higher current and also higher energy than the actual ones.

The intergroup reactors (their name in German is "Saugdrossel", which means "sucking reactor" and describes their function quite well) are connected so that their fluxes counteract and, if I remember correctly, results in their reactances cancelling when seen from the DC side. That would mean that they should not be included in rate of rise or final current calculation.

But, you have better check that out. Google "Intergroup Reactor" for more (and correct) information.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

OK, did some googling. IEEE has a good paper on this. But, since you need to be a member to read it, I just make this snippet available:

POZZOBON: TRANSIENT AND STEADY-STATE SHORT-CIRCUIT CURRENTS IN RECTIFIERS


"It has to be considered that the presence of the intergroup
reactor has no effect on the considerations made in this paper.
Being that the correct sharing of the dc current among the
two 6-pulse bridges of the 12-pulse unit is guaranteed by the
secondary reactances of the transformer only, which should
be equal, the intergroup reactor has no other feature than
supporting the instantaneous voltage difference between the
two bridges and reducing the circulation current to specified
value. The intergroup reactor has no dynamic effects on
the output dc current: the only smoothing effect on the dc
current can be associated to the leakage inductance due to the
unperfect coupling between the two parts of the winding of
the reactor. Such an effect could be requested as an additional
feature of the reactor to achieve a compact intergroup reactorsmoothing
reactor–capacitor assembly for the line low-pass
filter, but this case is not considered here. An ideal intergroup
reactor has no effect on steady-state and transient dc shortcircuit currents."

As said before, you do not need to consider the mid-tapped reactors in short circuit calculation.

There are lots and lots of rather heavy formulae in the paper. Easy to find, if you are an IEEE member.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[jimbofitz]

We are using a correction factor of approx. 1.35x VACrms. I was leaning towards the IPT's having little or no effect on the short but i'll have to read the article. This could be good news for our customer...

I just downloaded the paper and it looks heavy in equations for sure. I like the German translation, it is way better than the 10 names we use here in the US.

Thanks again for your help.

Regards,
Jim

 

[Skogsgurra]

Are you German? I have a very thorough paper on Sugdrosselschaltungen from 1967 (Gottfried Möltgen, Thyristoren in der technischen Anwenung). It has several pages with diagrams, text and equations, equations, equations. I can scan those pages if you are interested.


Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[jimbofitz]

Unfortunatly I cannot speak or read German.

 

[ShahAli]

I am working on a similar project. If you can upload that pdf for me i can get a translator to convert the pdf

 

[Skogsgurra]

Hello ShahAli! And welcome to the site.

I am not so sure that the average translator can do that succesfully. One of the scans is like this. The other eight are the same. But I can surely send them to you if you have a qualified translator and think it is wort while. I will need an e-mail address to send it to.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricuwe]

Gunnar,

instead of volunteering to translate I propose to post a reference in English tomorrow.

Jimbofitz,

you are correct, interphase reactors should have a small airgap, but this airgap is not intended to store energy - as it would be in an inductor.

The reason for providing this airgap is just to avoid saturation of the core in case of a small imbalance in current. Such imbalances allways can occure due several reasons.

Unsymmetrical fault current would immediately saturate a properly desiged interphase transformer leaving it just as an air-cored inductor in the circuit.

For the the border between inductor and transformer there is no clear definition. For me a device consisting of core and (one or more) winding(s) is an inductor if its purpose is to store energy in in a magnetic field. This leads to finding the correct formulas to design such devices.

All interphase transformers and coupled flyback inductors (usually called flybck transformers) I designed using this theorem proofed to work well.

 

[electricuwe]

The English textbook I had in mind is:

Bird, King, Pedder: An Introduction in Power Electronics, Wiley, 1993

But unfortunately it is by far not as detailed as the part from the Möltgen textbook.