Arc Flash Calculations

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derek22r

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How does cable distance between a fault and the protective device affect the incident energy level? I'm using software to do the calculations and in most cases, but not all, a longer distance results in a higher incident energy.
 
Perhaps you are finding that increasing the resistance reduces the short circuit current, thus increasing the time it takes the protective device to open. Does seem a bit counterintuitive though, doesn't it?
 
The incident energy is based on time and current. The longer cable increases the impedance, this will inherently decrease the fault current. But, by decreasing the fault current, you could then increase the amount of time to trip the breaker. therefore, in that case, the energy released during the event increases.

Hope that's what you where looking for.
 
jtester said:
Just wondering, if energy is proportional to I**2 x T and I goes down, why does T increase faster than I**2 decreases?

Jim T
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It has to do with the "speed" of the protective device which is not always strictly I^2 x t due to an intentional time-delay for most inrush applications.
 
I've seen examples where changing the bolted fault current by a 10% changes the trip time from milliseconds to seconds. In some cases, a 10% reduction in the bolted fault current might keep the breaker from tripping at all.

I do arc flash calculations at two points. I calculate the bolted fault current, and do one arc flash calculation. Then I do another arc flash calculation, usually using something like 80% of the the bolted fault current.

For both calculations, I look at the point where the arc flash current falls on the breaker curve and make sure they aren't on the verge of increasing the trip time a lot.

Steve
 
ramsy said:
Is this more effective, or cost effective, than using an arc fault breaker?
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The arc fault breakers are limited to low voltages, and low current applications.

Arc fault is always a concern, but at lower voltages the conequences diminish quite a bit. The arc fault calculations are more applicable at higher voltages were the breakers in question would most likely have adjustable trip settings, and be applied as large frame breakers, in MCC's, or utilize protective relays as in power circuit applications at 480V, or in medium voltage equipment.
 
If you analyze the ?root? incident energy equations (D.6.1 (a) and (b) in NFPA 70E) you will find there is indeed an ?I squared t? element and even a reductive I x t element; however, as Ron pointed out, both are generally overpowered by a linear element (t x Constant). ?I squared t? only works in the equation for an extremely short time where the fault impedance is assumed to be constant.

These equations are empirical in the first place so they are constantly subject to change as more test data are evaluated.

I cited the NFPA 70E equations because they are generally a bit easier for most folks to find than the IEEE 1584 material.
 
Roger,
An arc fault circuit breaker is somewhat unrelated to doing arc fault calculations.
Arc fault calcs are done to determine what type of protection an electrical worker needs when working on equipment energized.
An arc fault breaker detect an arc is to protect wires. There are some large breakers (>1600A) that are being called arc fault breakers manufactured by Siemens, but do not detect an arc, just open quickly at high fault currents with similarities to current limiting circuit breakers and fuses.
 
The longer the cable the more cable impedance which results in longer clearing times for the upstream protective device. The result may be higher energy levels. The PPE requirements may be higher in this case.
 
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