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| Arc Flash Boundary, Current-limiting Device https://brainfiller.com/arcflashforum/viewtopic.php?f=23&t=4303 |
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| Author: | GuillaumeRoux [ Wed Jan 18, 2017 2:13 pm ] |
| Post subject: | Arc Flash Boundary, Current-limiting Device |
Hey guys, My question is simple, I would like to know how to calculate the Arc flash Boundary for Current-limiting devices. For non-current limiting device, the equations are all listed in the IEEE 1584 standard and in the NFPA 70E handbook. There is a special section to find the energy from the specs of the current-limiting fuse but there is no mention on how to find the Arc Boundary of those fuses. Since the arcing time for the current limiting devices is around a quarter of a cycle (0.004 s), when I plug it in the non-current limiting device equations to isolate the distance, I get and arc boundary closer than the working distance which is set to 18 inch for the current-limiting fuse in their energy equation. I found a source where they give a table of the Incident energy and the Arc Boundary but I can't find a way to calculate the boundary. Here is the source, the table is found at the pages 10 and 11 : http://lightingcontrol.info/content/dam/public/bussmann/Electrical/Resources/technical-literature/bus-ele-an-3002-spd-electrical-safety.pdf Thank you for your time! PS : English is not my first language |
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| Author: | Jim Phillips (brainfiller) [ Thu Jan 19, 2017 10:46 am ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
A current limiting device begins to interrupt before the first quarter of an electrical cycle (before the first peak). It must clear the fault current before the first half cycle (zero crossing). This let thru energy can vary from device to device so what I have seen (by a software company) is use .01 seconds which is the bottom of the time current curve when the current is to the right of the fuse curve (current limiting region) - 0.01 seconds is the only value that is defined on the graph. You will likely have an arc flash boundary less than the working distance. This can also happen in some cases without current limiting devices. It just means the incident energy is quite low and you can (theoretically) get closer than the working distance before the energy of 1.2 cal/cm^2 is reached. The working distance is only the distance used to predict the incident energy reaching a workers torso and head. |
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| Author: | GuillaumeRoux [ Fri Jan 20, 2017 6:55 am ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
Thank you for your reply, I do understand the difference between the Arc-Flash boundary and the working distance. My problem is that the specifics equations for current-limiting fuses are only function of the available bolted three-phase short circuit current. For exemple, I have a class L fuses rated 1601 to 2000 A with an available bolted short circuit current of 50 kA, I use this equation found in the IEEE 1584 to get the incident energy in Cal/cm² : E = -0.1284 * Ibf + 32.262 which gives me about 25.84 cal/cm² ( Ibf is in Ka) Those equations are given with a working distance defined of 18inches. but I don't know the relation between the distance and the energy so I can't isolate the boundary when setting the energy at 1.2 cal/cm². but in the link I posted in my first post, they get a boundary with a current-limiting device. |
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| Author: | PaulEngr [ Sun Jan 22, 2017 1:23 am ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
The fuse equations as you have stated do not provide a way to calculate an arc flash boundary. This is only available from the empirical equation. However you can duplicate what the empirical equation does. The empirical equation proceeds in 4 steps: 1. Calculate the arcing current (we don't need this). 2. Calculate the normalized incident energy. 3. Calculate the denormalized incident energy taking into account the equipment design, working distance, and arcing time. 4. Alternatively, calculate the arc flash boundary taking into account the equipment design, desired denormalized incident energy, and arcing time. The denormalizing equations used in steps 3 and 4 are algebraically the exact same equation. The equation in step 4 is actually derived from step 3. This is important because it gives you the key to getting to the information you want. The normalized distance is 610 mm. There is a difference in the rate that the energy dissipates which depends on the equipment type which is the reason for the exponent X. Thus the distance term for scaling from normalized to denormalized incident energy is (610^X)/(D^X) where D is the working distance in mm. Let's call the working distance (not given in the post) WD and the new distance D. We can recalculate a different distance by multiplying by (WD^X)/(D^x). Simply plugging in different values for D will eventually reach the desired incident energy (normally 1.2 or 2.0) at different distances. X comes from the working distance table. You can also manipulate the equation to get an answer more directly like this: 1.2 = E*(WD^X)/(D^X) D = E*((WD^X)/1.2)^(1/X) Note that I'm not reinventing the wheel here. This is the exact same algebraic manipulation that is being done in the IEEE 1584 empirical equations except that in this case we are working with an incident energy value that has already been denormalized. Since we already have the calculation factors and arcing time "baked in", we can go directly to the distance factor. |
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| Author: | GuillaumeRoux [ Mon Jan 23, 2017 6:59 am ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
Thank you for your great answer it makes sense! |
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| Author: | jdsmith [ Mon Jan 23, 2017 8:49 am ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
One thing to think about - when a circuit is protected with current limiting fuses it is protected by three individual interrupters. For a three phase fault, one fuse will operate first, which changes the fault into a L-L or L-L-G fault, which will result in a different amount of fault current flowing in the connected phases. In some systems the amount of fault current flowing is now below the current limiting threshold, meaning the second and third fuses to operate will operate based on their TCC maximum clearing time, not the very short current-limiting time. If you calculate incident energy based on a 3 phase fault and you use a clearing time based on current limiting behavior, you may be underreporting the incident energy that would be experienced during a real world fault. If you want to be more accurate it is a good practice to use maximum clearing times from TCCs instead of current-limiting clearing times. |
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| Author: | PaulEngr [ Tue Jan 24, 2017 8:26 am ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
jdsmith wrote: One thing to think about - when a circuit is protected with current limiting fuses it is protected by three individual interrupters. For a three phase fault, one fuse will operate first, which changes the fault into a L-L or L-L-G fault, which will result in a different amount of fault current flowing in the connected phases. In some systems the amount of fault current flowing is now below the current limiting threshold, meaning the second and third fuses to operate will operate based on their TCC maximum clearing time, not the very short current-limiting time. If you calculate incident energy based on a 3 phase fault and you use a clearing time based on current limiting behavior, you may be underreporting the incident energy that would be experienced during a real world fault. If you want to be more accurate it is a good practice to use maximum clearing times from TCCs instead of current-limiting clearing times. IEEE 1584 contains several equations for specific types of fuses wherein they tested incident energy based on that type of equipment. Based on testing performed they gave empirically derived calculations for incident energy. Can you provide documentation for your theoretical and very complicated failure mode equivalent to the testing and documentation referenced by IEEE 1584? I mean it sounds good but the argument here is basically to ignore the testing and the standard and go out on a limb and try to determine incident energy based on a bunch of single phase stuff. Since we have no way to estimate single phase faults short of just assuming that they can't be any worse than a three phase fault, and your example has a bunch of strange fault currents that I really can't estimate (never mind clearing times), how can I actually model this? I'm at a loss as to how to estimate this strange L-L single phase or L-G-L two phase fault following a single or 3 phase L-G or L-L fault with a current limiting resistor, even if for a minute I can accept that it is possible and reasonable. I might as well go down the same rabbit hole with circuit breakers because for instance vacuum breakers don't all open at the same time...once the contacts are open even if they are tied to the same mechanical mechanism, they actually "open" at the next current zero crossing so all 3 phases actually stop conducting current at slightly different times (within about an 8 millisecond time period). |
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| Author: | Flash [ Tue Feb 14, 2017 9:52 pm ] |
| Post subject: | Re: Arc Flash Boundary, Current-limiting Device |
In general, if you are in the current-limiting range, you are below 1.2 cal/sq cm. One clarification, since the current peak shifts to approx .0083 sec during a fault (asymmetrical), the UL definition of current-limiting is a fuse that clears in less than .0083 seconds or before the peak. In reality the the Class j and RK1 fuses are probably clearing in less than 1/4 cycle, especially in smaller ampere ratings. As a rule of thumb fuses like the Mersen A4J and Bussmann JKS are going current-limiting at ten times their rating. |
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