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 Post subject: Short-Circuit problem
PostPosted: Tue Jan 24, 2012 10:48 am 
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I am performing an arc flash analysis for a new building. The short circuit specification suggests to calculate the maximum and minium interrupting duties (5 cycles to 2 seconds) short circuit currents.

I am not too sure how to move forward this. I have a momentary fault of 2.228kA at an angle of -87.19 degrees (this I am considering the maximum fault with all motors all, etc.). I also know the impedance to be .1515+j3.1056. My voltage is 12kV.

I am thinking this is a basic RL series circuit, but the math for me is not working out.

Can anyone walk me through on this?


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PostPosted: Tue Jan 24, 2012 3:17 pm 
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Motor contribution would be added for a maximum fault current, but not for a minimum.

What about the +/-7.5% manufacturing tolerance for transformer impedance?


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PostPosted: Tue Jan 24, 2012 5:11 pm 
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No I understand all of the variables regarding max and min. The problem I don't understand is how to calculate the fault current out to 2 seconds.


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PostPosted: Tue Jan 24, 2012 6:37 pm 
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bruinfan wrote:
No I understand all of the variables regarding max and min. The problem I don't understand is how to calculate the fault current out to 2 seconds.
Sorry quick response.

After a short time induction motors no longer contribute to a fault. Will other sources also be reduced or removed due to their protective devices operating, or other factors, prior to the 2sec point?

For example at 5 cycles you might use the X"d of a machine, but at 2 seconds you might be using X'd or even just X instead.

Do you have access to the IEEE Buff book?


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PostPosted: Tue Jan 24, 2012 8:57 pm 
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Yeah, I do have the buff book. I am looking at p.528 and it talks about interrupting duties out to 2 seconds and then refers to chapter 2 and says that these values are obtained in it. I went through chapter 2 and did see anything that might help.

Good question. I am not sure really. The specification doesn't specify what they want so I don't know exactly.

What do you think?


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PostPosted: Wed Jan 25, 2012 6:06 am 
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Without knowing anything about your protective devices or the one-line, I would not be able to tell you for sure what your customer is looking for. It is possible that they have simply found a multi-page specification, and are using it because 'more detailed, must be better'.


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PostPosted: Wed Jan 25, 2012 6:11 am 
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That thought actually crossed my mind as well. Thanks for all your input on this.


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PostPosted: Fri Jan 27, 2012 1:02 pm 
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JBD wrote:
Sorry quick response.

After a short time induction motors no longer contribute to a fault. Will other sources also be reduced or removed due to their protective devices operating, or other factors, prior to the 2sec point?

For example at 5 cycles you might use the X"d of a machine, but at 2 seconds you might be using X'd or even just X instead.

Do you have access to the IEEE Buff book?


bruinfan wrote:
I am performing an arc flash analysis for a new building. The short circuit specification suggests to calculate the maximum and minium interrupting duties (5 cycles to 2 seconds) short circuit currents.



I am not too sure how to move forward this. I have a momentary fault of 2.228kA at an angle of -87.19 degrees (this I am considering the maximum fault with all motors all, etc.). I also know the impedance to be .1515+j3.1056. My voltage is 12kV.



I am thinking this is a basic RL series circuit, but the math for me is not working out.

Can anyone walk me through on this?


[SIZE=3][SIZE=2]I'm not sure about the 2 second part but here is the short circuit current calculation based on your data:[/size][/size]

Impedance: 0.1515 +j3.1056 Ohms

X/R ratio = 3.1056 / 0.1515 = 20.5
Angle = inverse Tangent (20.5) = 87.2 degrees

Short circuit current = voltage line-neutral / impedance magnitude

Impedance magnitude = square root (0.1515^2 + 3.1056^2)
3.109 ohms

Voltage line-neutral = 12000/sqrt(3)
= 6928 volts

Short circuit current = 6928 / 3.109 = 2228.4 Amps

You can have momentary and interrupting ratings which are several cycles apart but I have only heard of 2 seconds being used as an arc flash time cut off. I wonder if this is what they meant and mixed it up?

_________________
Jim Phillips, P.E.
Brainfiller.com


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PostPosted: Thu Feb 02, 2012 8:26 am 
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Which Software are you using? for arc flash. In ETAP you have to specify or it automatically calculates for required no. of cycles. You also have to mention the fault clearing time in ETAP settings..


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PostPosted: Fri Feb 03, 2012 1:39 pm 
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It really isn't that complicated.
First determine what your starting short circuit current is.
Then look at the operating time all of your protective devices that are seeing this current.
As time goes by, different protective devices operate, thereby removing some sources of fault current.
After 2secs of time, determine which sources are still active, and how much current they are supplying into the fault.
This is your 2sec value.

You could use this value to determine if the remaining protective devices still 'see' enough current to operate.
You could use it to determine what forces your conductors and equipment will be exposed to (although the 30 cycle value is more typical)
Or, as Brainfiller pointed out, you can use it in arc flash calculations. Hmm, maybe the 2sec 'arc flash cut-off time' came from the IEEE Buff Book?


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