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 Post subject: Max short-circuit current used for Arc-Flash calculations
PostPosted: Wed Apr 25, 2012 12:48 am 

Joined: Sun Jul 22, 2007 5:00 pm
Posts: 32
Location: Stratham, SW Australia
IEEE 1584 Section 4.3 Step 2 requires arc-flash hazard calcualtions for both max and min short-circuit currents.
Is it necessary to include stand-by generator contribution is cases where the stand-by generators only run for three hours per year ?


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PostPosted: Wed Apr 25, 2012 3:38 am 
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It is quite frequent to have multiple labels for different scenarios when generators are in use or when you have double ended substations with three conditions (feeder A, feeder B, both feeders). Once you get somewhat downstream from the generator, the arc flash values begin to be influenced by the upstream device only.

In these scenarios, the alternative method is to look at the worst case value only.

Another alternative methodology would be to use fault trees to look at the exposures because the already low probability of failure considering the rate of exposure to the hazard (3 / (365 * 24)) is so low that as you said, it may not be worth considering. However I would caution you that if the 3 hours of exposure are specifically because of tasks that are done only while the generator is operating, then you should take into consideration the exposure during those tasks (which is ALWAYS).

Note that generators are particularly troublesome for incident energy calculations because typically the impedance and available fault currents are REALLY high. I prefer to have generator-mounted circuit breakers/fuses for that reason and to have task instructions that involve shutting down the generator, and have series circuit breakers (if I need it) rather than touching the on board circuit breaker which typically has an excessively high incident energy value rating.


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PostPosted: Wed Apr 25, 2012 10:20 am 
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You're potentially looking at three calculations: minimum utility, maximum utility (emergency generator in parallel) and emergency only. The contribution from a small emergency generator running in parallel to the grid is not likely to have a big impact on hazard levels. Unless you have specific instructions that prohibit work during parallel generator operation I would just include it in the maximum case.

Paul already already covered some of the issues with emergency generator operation and work like fault finding does happen during unplanned outages. You definitely need a separate calculation for this case, as fault levels and trip times can be very different from normal conditions.


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PostPosted: Mon Apr 30, 2012 8:31 am 
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You absolutely must run a stand-by generator scenario and determine the arc flash energy. Most protective devices have what is called an inverse characteristic. This means that the more fault current that flows through a protective device, the faster it will trip. Less fault current will cause the protective device to trip slower. When a generator feeds the equipment, usually the fault current is lower than the utility. This is because the generator has a larger impedance and therefore is a smaller source of fault current.

Time tends to impact the arc fault calculations greater than the amount of fault current. However, the amount of fault current can change the protective devices operating times. Therefore, it is very important to run a scenario with the stand-by generator on line. Then compare the arc flash energy to the other operating scenarios and select the scenario with the highest energy level. This is the value that should go onto the arc flash level.

_________________
Robert Fuhr, P.E.; P.Eng.
PowerStudies


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PostPosted: Fri Jun 29, 2012 8:35 am 

Joined: Fri Jun 22, 2012 6:59 am
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I agree with Robert. It is easy to believe that the higher the fault current, the higher the incident energy. However, as robert mentioned, tripping time is a large factor that must be considered. As fault current decreases tripping time can increase, resulting in potentially longer arc flash exposure and higher incident energy.

E=4.184 CfEn(t/0.2) ((610^x)/(D^x)

E=incident energy
t= arcing time (based on protection tripping time)

They say to use 2s max if resulting time is longer due to the high possability that the person will be out of range after the 2s have passed.

If anyone would like to add to this, I am open.

C


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