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 Post subject: Clearing Time
PostPosted: Wed May 12, 2010 4:12 pm 

Joined: Mon Mar 29, 2010 8:31 am
Posts: 24
Location: Jonesboro, AR
Ok, been through the book and through my sheets for fuses. I need some help in determining the proper protective device clearing time.

System
Main Transformer = 13.7V ~ 480V 1.5MVA %Z=5.67


I don't have information on the 3000A fuses on the main switch gear (will get this as we do Thermography in the next couple weeks)

MCC section = Fused off main switch gear 1200A KRP-C-1200 Class L time delay.

I(sc) is 32kA (I rounded up)
Arcing Current is calculated at 17.69kA

Looking at the fuse time-current curves the 1200A line runs a of 0.08 or 0.09 seconds. Since this is melting time chart I found "If they show only the average melt time, add to that time 15%, up to 0.03 seconds, and 10% above 0.03 seconds", NFPA-2009 also says that additional time must be added if a time delay function is installed and operating, but no reference on the amount of time. How much extra time? Is there a standard, or am I missing something else?


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PostPosted: Mon May 17, 2010 6:23 am 
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Location: Cincinnat, OH
First of all, are you performing arc flash study? If you are then you are using the wrong SC value.....32kA is an infinite source calculation. For accurate results you should know what the Utility values are and use them in conjunction with the transformer. I think you are goiong to find that the arcing current in calculation will be much lower, hence prompting a longer clearing time for your fuses. The Current would probably be much less by the time you get to the MCC because of cable impedance as well.

Are you just trying to coordinate?


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PostPosted: Wed May 19, 2010 7:45 am 

Joined: Mon Mar 29, 2010 8:31 am
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Location: Jonesboro, AR
I(sc) calculated based on NFPA 70E Annex D equation [D.2(a)]

I(sc) = {[MVAx10^6]/[1.732*V)]}*(100/%Z)

From my previous post

I(sc) = (1500000/831.36)*17.636684
I(sc) = 31821A or 32kA

Secondary I(sc) from transformer calculations
I(secondary side) = KVA*1000/(V*1.73) = 1500*1000/(480*1.73) = 1806A
I(sc) = I/%Z = 1806/0.0565 = 31970A

So, I'm pretty comfortable with the 32kA value. From the calculation, I know that this is the max short circuit amps for the system. Using the Annex D and equation [D.7.2(a)] I calculated the arching current at the switch gear using a gap of 32 from table D.7.2 and got 16.57kA. Which is the same as some of the online calculators have given me like:

http://www.arcadvisor.com/arcflash/ieee1584.html

So this is the arcing current that can enter the switch gear. The gear is fused by a 3000A fuse (that's all I know about it), so I can't calculate an acceptable flash boundary for this piece of equipement until I know the device opening time is (or use the standard 0.2s [as equation D.7.3(a) does], but the with a time delay fuse this could be longer, or shorter if it's current limiting), but I do know the fused discconnects for the first 4 MCC sections. All 1200A KRP-C-1200 Class L fuses.

Redoing the above arching current for an MCC section (gap changed to 25) the arcing current becomes 17.69kA. The average melt chart shows fuse breakdown in 0.02s to 0.03s (its log scale and a pain in the ass to read/guesstimate). Apparently I read it wrong must have been looking at 1600A fuse. Add 15% for melt time chart times up to 0.03s (found this on the arc flas site as well). Device opening time = 0.023s, at 85% of arc current (15.95kA) time from chart is still 0.03s.

The NFPA book discusses the opening time of a breaker with a saftey rating of 2 cycles, but this is a fuse and a time delay fuse at that, so if the arc is something less that 17 kA the system could very well have a longer arc event.

Distance from the main transformer is less than 50ft, so any voltage drop at this point of fault would almost be negliable (25' of 500kcmil, 6 per phase service is reduces the I(sc) by less than 2kA).

I'm pretty comfortable with my numbers, but I'm not comfortable with the time delay fuse and the NFPA book saying that "Additional time must be added if a time delay function is installed and operating", but no example of how much time and a chart showing 600V line side opening times of 60-120 cycles.


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PostPosted: Wed May 19, 2010 7:59 am 

Joined: Mon Mar 29, 2010 8:31 am
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Location: Jonesboro, AR
Sorry didn't finish my thought on:

Distance from the main transformer is less than 50ft, so any voltage drop at this point of fault would almost be negliable (25' of 500kcmil, 6 per phase service is reduces the I(sc) by less than 2kA).

While this might raise my opening times, this isn't where my issue is.


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PostPosted: Wed May 19, 2010 10:40 am 
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I am not sure what you are trying to calculate here, AF boundary or Incident Energy? Sounds like you are mixing methods. NFPA 70E methods based upon ralph lee equations don't account for arcing current. The method you referenced in d7.2 deals with IEEE 1584 methodologies. I would either stick to one or the other. If you are going to go the IEEE, which I would recommend, than you should use d7.5 to calculate the AF boundary not d2a.


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PostPosted: Wed May 19, 2010 1:02 pm 

Joined: Mon Mar 29, 2010 8:31 am
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Location: Jonesboro, AR
Yes, I'd like to use the IEEE calculations

Still need I(bf) - bolted fault (or short circuit, yes?), still calculated off the transformer with the short circuit contributions of the motors and wire.

I'm calculating the fault current after the feed to one MCC panel (holding the motor and resistance to nominal, and I understand this only gets me close)

Based on the above with a Fault at a point after the fused 1200A disconnect, the maximum short circuit current I could ever get at that location is 32kA. Based straight off transformer short circuit analysis.

So the bolted fault at that point is equal to the full load of the transformer at that point until the fuse pops (correct? ignoring motor load and wire contributions). So if I could take a wire and a meter and short the secondary of the transformer I would get 32kA to read on the meter.
I(sc rms)= I(sc)*M where I'm holding M to 1.

Using equation D.7.2(a) with I(bf) = 32kA gives an ArcCurrent of 16.57 kA.
And the system falls into the system limits outlined in D.7.1

Calculating the Incident Energy using D.7.3(c)
t = ArcingTime = Time until fuse opens at the calculated ArcCurrent.
Now the fuse is a time delay fuse, opens at 0.03s for an ArcCurrent of 16.57 kA. On page 70E-63 left hand column it states that "additional time must be added if a time delay function is installed and operating".

My question is:
1. Is there a standard amount of time to add to the ArcTime? like 2 cycles for breakers.
2. If not, do I use the standard 0.2s arc time the equation in D.7.3 is based on to calculate my Incident Energy.
3. What happens if the arc current is not big enough to pop the fuse in 0.2s and it runs on for 60 cycles before breakdown, do I even need to worry about this?

I want to see if I fall within the notes of table 130.7(C)(9).

I do not have all the information to include wire size, number and C values. I'll gather these as we do maintenance on the system and re-calculate.


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PostPosted: Thu May 20, 2010 5:55 am 

Joined: Mon Mar 29, 2010 8:31 am
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Location: Jonesboro, AR
Thought about this some more, and I got caught up in the details instead of my question.

It doesn't really matter what the bolted fault current is for my question. Weather it's measured at 32kA or 6kA or 1200A. The fuse is a time delay fuse, and determining the clearing time of the fuse is the issue, or use of the proper time to calculate IE what's accepted what the standard is.

The exercise was just a crank on the fomulas to see what falls out assuming no loading and no resistance in the wires. Just a basic loop of the transformer and the fuse.

I did a pretty good job of mixing what I was doing with where I'd like to be, and that isn't the case, I apologize for throwing anyone for a loop.


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PostPosted: Thu May 20, 2010 11:18 am 
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Joined: Wed Jun 04, 2008 9:17 am
Posts: 428
Location: Spartanburg, South Carolina
To get the clearing time, you need to know the portion of the arcing current flowing through the fuse and look at the Time-Current Curves (TCC). Using the maximum transformer fault current may result in lower IE because the clearing time would be faster. This is why you need to look at minimum fault current as well as maximum. The IE will also be affected by motor fault contribution. Note that the motor contribution does not flow through the fuse.


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PostPosted: Thu May 20, 2010 3:47 pm 

Joined: Mon Mar 29, 2010 8:31 am
Posts: 24
Location: Jonesboro, AR
One sentence in this book is what's throwing me.

For breakers (5kV to 15kV) clearing time is approx. 0.1 ~ 6 cycles.

Broken down:
Actual Breaking Time ~ 2 cycles
Relay Operation ~ 1.74 cycles
Safety ~ 2 cycles

Additional time must be added for a time delay function

I look at the fuse that's being used. It says "Time Delay", I look up the clearing time on the TCC and it comes out to X. Do I need to add more time to X based on the statement above? If so, how much? Or why not.


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PostPosted: Thu May 20, 2010 4:38 pm 
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What book are you referring you and where? Breakers built to the preferred ratings of IEEE C37.06 have interrupting times of 3 or 5 cycles.

But why would you use breaker and relay time for a fuse? Use the fuse clearing time.


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PostPosted: Thu May 20, 2010 6:32 pm 
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Joined: Wed Jun 04, 2008 9:17 am
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Location: Spartanburg, South Carolina
Fuses have a minimum melting time and a total clearing time. Use the total clearing time.


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