Short Circuit Current contribution.

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Besoeker

Besoeker

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We have a 6,300 kW wound rotor induction motor.
The stator voltage is 11 kV, 50 Hz, the rotor 3.2 kV
It is run up to about 1000 rpm using a stepped resistance starter.
Then it changes over to variable speed control over the range 1000 rpm to about 1485 rpm.
It drives a compressor.

The speed controller comprises a rotor rectifier, a step up chopper to about 950 Vdc, then through a mains commutated inverter into a 950 Vac recovery transformer which steps it up to the 11 kV supply voltage. Sounds complicated but it is done in the interests of efficiency - around 96% efficiency including the drive and motor. And we had to prove it.

But my question is very simple. If there was a short on the 11 kV bars, how much fault current would this arrangement add?
 
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GoldDigger

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That sounds like the kind of question I would look to member Besoeker to answer. :angel:

More seriously, I do not think that the rotor controlled rectifier side will be capable of any power feedback, leaving the mains-commutated (grid interactive?) inverter.
Leaving out completely the available kinetic energy of the motor system, I would look at what the output current limit of the inverter is before either damage or protective shutdown with a dead short directly at its output. Then reduce even that based on the percent impedance of the coupling transformer.
 
mayanees

mayanees

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Westminster, MD
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Electrical Engineer and Master Electrician
I'd say about 8 times its running fla, or 2,645 amps for a controller-bypassed spinning load contribution.
 
Besoeker

Besoeker

Senior Member
Location
UK
GoldDigger said:
That sounds like the kind of question I would look to member Besoeker to answer. :angel:
Click to expand...
Droll..........)

GoldDigger said:
More seriously, I do not think that the rotor controlled rectifier side will be capable of any power feedback, leaving the mains-commutated (grid interactive?) inverter.
Leaving out completely the available kinetic energy of the motor system, I would look at what the output current limit of the inverter is before either damage or protective shutdown with a dead short directly at its output. Then reduce even that based on the percent impedance of the coupling transformer.
Click to expand...
A couple or three points.
The rotor rectifier is a plain, uncontrolled rectifier. It is the first stage and source of the power feedback system. The step up chopper is the next, and the final stage is the mains-commutated inverter. There high speed semiconductor fuses at the output of that inverter.

If the mods will allow it, I will post a link to our brochure. I don't think it would breach any confidentiality since that division has more or less gone since I and a few others left the organisation.
 
Jraef

Jraef

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So you are describing a Slip Energy Recovery system on a WRIM then? In the SER system I had worked on, the recovery inverter still functions to contribute to the available fault current if the motor is spinning. But I've never been able to get a decent answer as to how much energy is involved. From what I gathered when I tried to research this a few years (decades?) ago, the recovered ENERGY varies as the inverse of the speed change, so in theory then the amount of fault current contribution would follow somewhat, limited and changed by the transformer impedance. The math on that would make my head hurt though, so sorry but I can't really contribute.

Where I got the backward inclined energy recovery data from though was an old paper I found back when I had to do this. I was able to dig out the paper copy I had and having the title and the authors, I was able to find a copy on line. its from 1992 and is discussing GTO based SERS drive systems, but you might still find it useful. Most of the other stuff I found was way too vague when it came to actually quantifying the recovered energy, they just kept discussing that it DID recover it and left it at that.
http://users.encs.concordia.ca/~pillay/c88.pdf
 
Besoeker

Besoeker

Senior Member
Location
UK
Jraef said:
So you are describing a Slip Energy Recovery system on a WRIM then? In the SER system I had worked on, the recovery inverter still functions to contribute to the available fault current if the motor is spinning. But I've never been able to get a decent answer as to how much energy is involved. From what I gathered when I tried to research this a few years (decades?) ago, the recovered ENERGY varies as the inverse of the speed change, so in theory then the amount of fault current contribution would follow somewhat, limited and changed by the transformer impedance. The math on that would make my head hurt though, so sorry but I can't really contribute.

Where I got the backward inclined energy recovery data from though was an old paper I found back when I had to do this. I was able to dig out the paper copy I had and having the title and the authors, I was able to find a copy on line. its from 1992 and is discussing GTO based SERS drive systems, but you might still find it useful. Most of the other stuff I found was way too vague when it came to actually quantifying the recovered energy, they just kept discussing that it DID recover it and left it at that.
http://users.encs.concordia.ca/~pillay/c88.pdf
Click to expand...

Good paper. The Bashir Al Zahawi mentioned a couple of times in the references worked for me for some years. Actually lodged with my family when he first came to work for us.
We worked together on the very drive in my opening post. We called these drives the ISK for Improved Static Kramer. The step up chopper means there is no circulating current at top speed hence the high efficiency and much better power factor than the conventional SER system. We supplied these for large variable speed drives on pumps and compressors mostly in the water and petrochem industries.

I can post a link to the brochure if mods here will permit me to do so.
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Besoeker...

It would add a little, about 4%! The electronics "package" recovers 2x rated slip-loss, which, for example, is 1%! If you ignored other system/component losses, then, fault-current contribution is 4%! More, if you were able to reduce the motor's transient-reactance!

Phil
 
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Besoeker

Besoeker

Senior Member
Location
UK
Phil Corso said:
Besoeker...

It would add a little, about 4%! The electronics "package" recovers 2x rated slip-loss, which, for example, is 1%! If you ignored other system/component losses, then, fault-current contribution is 4%! More, if you were able to reduce the motor's transient-reactance!

Phil
Click to expand...
It actually recovers about 15%.
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Besoeker...

You are to be lauded if you're part of the design team!! Frankly, I'm more interested in knowing what characteristics were altered, and resultant recovery power and Pf, achieved!

Phil
 
Besoeker

Besoeker

Senior Member
Location
UK
Phil Corso said:
Besoeker...

You are to be lauded if you're part of the design team!! Frankly, I'm more interested in knowing what characteristics were altered, and resultant recovery power and Pf, achieved!

Phil
Click to expand...
Thank you. It's what I did for a living - designed power electronics.
The significant difference from the conventional Kramer is the step up chopper. We ran the recovery bridge at a fixed firing angle.
 
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Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Besoeker...

Thus, a key was insertion of the VFD... But I'd still like to know about the increase of restored power, especially the reactive component, above conventional design!

Also, does your company participate in the American market?

Phil
 
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