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I have a setup where threre is a 6 pole 3 phase rectifier connected to a 690VAC bus. How do I calculate the AC fault levels for a fault on the DC side? The rectifier is used to supply power for DC Variable speed drives.
The lecturer said to look into "Effective commutation" and also dc volta= ac volts/pi.
But I cant find anything much on google to get started with this. Also searching for the things he said didn't come up with anything useful.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
In order to calculate the AC fault level, you'd have to know the DC fault level. Ask your lecturer how he would calculate that. If he mentions MVA, ask for a rebate on your tuition fee and look for another university.
If it were me, I'd assume a dead short on the DC bus. This equates to a dead short between all three AC inputs of the rectifier, and you can calculate the AC fault level in the normal way for a 3 phase fault.
If the DC fault doesn't drag the bus voltage all the way to zero, it gets really complicated. You have to find the actual bus voltage, the power delivered to the fault (MW, not MVA) and then calculate the power factor of the rectifier under those conditions, to get the AC fault level. This will involve dealing with overlap caused by the reactance of the AC line, which I think is what the guy meant by effective commutation.
In order to calculate the AC fault level, you'd have to know the DC fault level. Ask your lecturer how he would calculate that. If he mentions MVA, ask for a rebate on your tuition fee and look for another university.
If it were me, I'd assume a dead short on the DC bus. This equates to a dead short between all three AC inputs of the rectifier, and you can calculate the AC fault level in the normal way for a 3 phase fault.
If the DC fault doesn't drag the bus voltage all the way to zero, it gets really complicated. You have to find the actual bus voltage, the power delivered to the fault (MW, not MVA) and then calculate the power factor of the rectifier under those conditions, to get the AC fault level. This will involve dealing with overlap caused by the reactance of the AC line, which I think is what the guy meant by effective commutation.
thanks Steve. Any idea where the dcvolts=acvolts/pi comes in?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
If there were an inductor in the DC link, then that equation would hold true. The inductor averages the voltage, and the relationship between average and peak of a rectified sine wave always has a factor of pi in it.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
In a real-world situation the fault current would cause the 3-phase rectifier to short-circuit itself presenting a dead-short to the ac, then either part of the rectifier assembly will explode or some part of the ac circuit will blow open (wire, fuse(s), circuit breakers etc.)
Wikipedia is your friend
There you will see where the PI comes from, but that will not really apply as 'average' dc voltage is not relevant.
For a variable speed dc motor the traditional method would be to use 3 diodes and 3 thyristors as the rectifier using phase-angle control (like in a light dimmer) to adjust the speed, possibly with field-weakening. Briefly GTO thyristors and now IGBTs are used to 'chop' the ac rectification at kHz rates using the inductance of the armature to 'smooth' the current waveform, which makes calculations a little more complex.
Of course in the real world there should be a 'semiconductor' fuse in each of the 3-phase lines. . These fuses aren't cheap, e.g. I replaced a 160A fuse yesterday, £27 + tax due to a fault on the output of an 11 kW drive (inverter for ac motor). The fuse was between 3-phase rectifier/Bus-caps and the igbt output bridge, which it protected.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Sulaiman you are absolutely right! In a real world situation you would assume the rectifier shorts out completely, and simply use the fault level of the mains supply to calculate the interrupting ratings for those "semiconductor" fuses. You have to be conservative, because if a fuse gets hit with a fault too big for it to clear, it can explode, start a fire, people can get hurt and so on. Plus it makes the calculations easier!
But this is one of these sadistic problems I remember from my power electronics classes.
I suppose the problem could be restated: Knowing the impedance of the AC supply, what DC current could you expect if you shorted the rectifier output? This isn't of any practical use because the rectifier is always fused on its AC inputs, but it might help to visualise the problem.
Registered Member #2431
Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
Wikipedia is your enemy.
Or so I have been taught at every college Ive ever been at.
I concur with those who advocate for the dead short assumption for this problem, but the professor should have specified this if he meant for it to be otherwise.
If he did mean for you to solve this at some mid-point value, then your professor is evil. Enjoy the math.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
Yes, I too went through the academic torture mill ! (B.Sc. Electrical & Electronic Engineering, Birmingham, 1974) ..yes I know, most 4HV members weren't even born then !
Teaching in such a manner is like IT students learning Pascal; an interesting/useful academic exercise but ultimately not actually applicable. I had two types of lecturer, those that had never left academia and the good ones! Unfortunately to get a degree one needs to adopt the academic approach only. How nice engineering would be if in the real world only ideal situations existed !
So, the dc side will 'look' like a resistance (fault) and inductance in series if you can assume that the resistance is negligible and the inductance can be ignored (except during the initial transient time) you are left with 3 phases, each with their source impedances in series with a pair of inverse-parallel diodes connected to a common 'star' point and as the diode drops are negligible compared to 690/sqrt(3) volts you are left with a simple ( ?? ) ac fault situation.
If commutation enters into the equation then the assumption of negligible fault resistance has to be dropped and I would like to see the answer because I've grown too lazy to work out such things.
So, scrooch, please ignore my ramblings but post the 'correct' answer here to put me out of my misery !
Registered Member #1334
Joined: Tue Feb 19 2008, 04:37PM
Location: Nr. London, UK
Posts: 615
Sulaiman wrote ...
Yes, I too went through the academic torture mill ! (B.Sc. Electrical & Electronic Engineering, Birmingham, 1974) ..yes I know, most 4HV members weren't even born then !
BSc Electrical & Electronic Engineering, Southampton, 1980. Recently I found my uni notes in the loft.
Grads? Divs? Curls?
Couldn't understand a word! - nor do I want to - never used them, never will...
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DC fault levels and AC fault levels
