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 Post subject: Equipment Evaluation and Interrupting Rating Regarding Asymetrical and Symmetrical Fault Current
PostPosted: Wed Nov 13, 2013 12:58 pm 

Joined: Tue Mar 12, 2013 12:31 pm
Posts: 14
Location: Dallas, TX
Forgive me if this has been previously discussed but I couldn't find any threads regarding this specific encounter.

I am failing some OCPD in equipment evaluations due to X/R system > X/R test, thus increasing Calc Int kA. I did some research to wrap my head around what was actually happening in SKM and from what I gather to put it shortly.. SKM has a judgement fail (*N1) that if my network has a system X/R greater than what the OCPD tested X/R is then it evaluates the bus at the more conservative X/R. Now I have two questions that people seem to get frustrated trying to explain or explanations vary. 1) X/R has to do with the exponential time constant (X/2*pi*f*R) from DC offset, right? So if the network is highly Inductive, this yields a higher X/R, which in turn increases the time constant and DC Offset? So if the DC offset is added to the symmetrical fault value that is what yields the Asymmetrical signal?

2) Assuming that above is right, some research I have found regarding OCPD interrupting rating is that if the X/R is evaluated higher than tested, then the equipment could be underrated. However, it is my understanding that in the asymmetrical portion of the wave, the time period is so small and happens so fast that by the time the breaker senses the inflated asymmetrical fault current the asymmetrical portion has already completed and the signal has long been at the symmetrical fault magnitude by the 4-6 cycle time. A coworker stated that even for the worst case scenario that if the fault happened on the wave where Voltage is at 0 and this yields a large asymmetrical fault current, that we still only worry about the stable symmetrical value because the energy of the higher current doesn't have time to heat up because it decays so fast. If this is the case, then why is the equipment evaluation saying the OCPD is underrated.

Any comments on errors in my thought process are greatly appreciated.


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PostPosted: Thu Nov 14, 2013 11:50 am 
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Josh Gatlin wrote:
... the signal has long been at the symmetrical fault magnitude by the 4-6 cycle time....


Most molded case breakers, built to UL489, have instantaneous tripping functions so they clear faults within 3 cycles. Those that have not cleared the fault are have probably already had theirt contacts begin opening.

We always consider the X/R ratio, and adjustment factors, when evaluating molded case breakers.


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PostPosted: Thu Nov 14, 2013 12:55 pm 
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The rate of asymmetrical signal decay is determined by system time constant L/R (X = w x L) at point of fault. I doubt that
Quote:
the time period is so small and happens so fast that by the time the breaker senses the inflated asymmetrical fault current the asymmetrical portion has already completed and the signal has long been at the symmetrical fault magnitude by the 4-6 cycle time

In fact, I believe the opposite is quite true. As an example, consider system with X/R equal 15. It corresponds to time constant of approx 0.05sec on 50Hz network when it takes around 0.25sec or 15 cycles for the assymetrical component to decay. I believe your equipment evaluation is right stating that the OCPD is underrated. You brought up a good point emphasizing how critical is to factor-in protection device X/R rating (along with the device I.R. rating) when selecting the OCPD.

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PostPosted: Fri Nov 15, 2013 8:48 am 

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Thanks for the responses! Now, I have two follow up questions. 1) Where can I find a reputable study on low voltage circuit breakers and associated clearing times. (NOT opening times). Some peers are disputing the
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Most molded case breakers, built to UL489, have instantaneous tripping functions so they clear faults within 3 cycles.


2) Also I would like to know the source where I can find this derivation. Thanks!!
Quote:
It corresponds to time constant of approx 0.05sec on 50Hz network when it takes around 0.25sec or 15 cycles for the assymetrical component to decay.


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PostPosted: Fri Nov 15, 2013 11:32 am 
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Quote:
2) Also I would like to know the source where I can find this derivation. Thanks!!

The asymmetrical component declines exponentially, and after one time constant period will have reached 36.8% of the initial state value. It is considered to reach zero in five time constant periods. Assuming X/R = 15, since X = w * L = 2 * pi * 50HZ * L ~ 314 * L, time constant equals L/R = (X/R) / 314 = 15/ 314 = 0.05sec and it takes roughly 0.05 * 5 = 0.25sec or 0.25/0.016 = 15 cycles for the inductance to discharge and fault current to stabilize.

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PostPosted: Fri Nov 15, 2013 11:48 am 
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Josh Gatlin wrote:
Thanks for the responses! Now, I have two follow up questions. 1) Where can I find a reputable study on low voltage circuit breakers and associated clearing times. (NOT opening times).


How about a paper from Eaton?

http://www.ewh.ieee.org/cmte/ias-esw/Abstracts/ESW2009-18%20Application%20and%20Maint%20of%20MCCBs%20Rev2.pdf

[SIZE=12px]"Don’t Circuit Breakers Take Longer To Clear A Fault Than Other Types Of Overcurrent Protective Devices?[/size]
[SIZE=12px]In many cases, no! Most real world faults are lower level, single phase arcing faults. In these situations, MCCBs will often clear the fault faster than other types of overcurrent protective devices. When operating in their instantaneous overcurrent range, the majority of today’s MCCBs will clear a[/size]
fault in less than one cycle."

Do your peers understand that breaker TCCs do show clearing times?


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