how is voltage related to arcflash hazard analysis?

[learning99]

With only an electronics background, my electrical knowledge is small... but I want to learn... My understanding is that bolted fault current and over current protection clearing time are the 2 factors involved in determining arcflash hazard. I continually see, hear folks talking about voltage. How or is it related? Would power distribtution with x current, and y clearing time present the same hazard at 240Vac, or 480Vac single phase? Many thanks. Learning99

 

[ron]

The voltage is included in the calculation because it effects the ability for the arc to be sustained.

 

[zog]

With only an electronics background, my electrical knowledge is small... but I want to learn... My understanding is that bolted fault current and over current protection clearing time are the 2 factors involved in determining arcflash hazard. I continually see, hear folks talking about voltage. How or is it related? Would power distribtution with x current, and y clearing time present the same hazard at 240Vac, or 480Vac single phase? Many thanks. Learning99

As Ron mentioned, self sustaining is the key. here are many factors but voltage is not something you can really use to judeg the hazard level, usually you will have a larger arc flash hazard from a 480V system than you will in a 15kV system.

 

[charlie b]

My understanding is that bolted fault current and over current protection clearing time are the 2 factors involved in determining arcflash hazard.

Not quite. It is not about the bolted fault current. That event will cause a rapid tripping of the breaker, which will terminate the arc flash. A far lower current will be allowed to flow for a much longer time, before the breaker trips, and that can result in a higher total arc flash energy.

 

[learning99]

Thank you for your quick responses: May I ask if you could elaborate a little further on why 480V would sustain LARGER arc that 15kV? Seems like large voltage would give more energy to sustain. Again, thanks for your help. Learning99

 

[charlie b]

Seems like large voltage would give more energy to sustain.

It would also cause the breaker to trip sooner. Time is an important parameter in the total amount of arc flash energy that will be released.

 

[learning99]

Thanks!!

 

[zog]

It all depends on the OCPD. You have to look at a wide range of fault currents, an arcing current value can vary widely as compared to bolted faults. When an analysis is done several different possible arcing current levels are looked at and how long they would arc depends on the OCPD's time current curve. Breakers can have different trip functions and settings. When the device clears the fault is a huge factor and as Charlie mentioned often it is the lower current arcing faults that create the larger hazard. You are required to do at least the lowest and highest possible arcing currents as part of your analysis, software like SKM does many, many different points. The worst case senario is what goes on the label.

A common problem is LV switchgear where the Main that protects the gear does not have INST trips (for coordination reasons), in this case even the highest current arcing fault will not be cleared right away, so the STD comes into play. It is very common to see 480 subs with >40cal/cm2.

A good way to think about it is waving your hand over a heat source. A big fire will not burn your hand if you pass it over it very fast, hold you hand over a candle for 5 seconds and you get a serious burn.

 

[gar]

110218-1427 EST

An arc discharge is primarily a low voltage load with the current thru it largely determined by the external circuit.

From "Fundamentals of Engineering Electronics", by William G. Dow. p456, and 552.

An approximate empirical formula for an arc voltage drop is:

Va = A + B / Ia^N

where A, B, and N are empirically determined constants.
Va is the arc drop, including both electrodes, and
Ia is the arc current.
N is a function of the anode material.

If you have a constant source impedance at say 100 ohms at 10,000 V, then the short circuit current is 100 A, and assume 50 V for the arc drop. Then the power dissipated in the arc is 5000 W.

Use a transformer to step this down to 100 V, then the approximate source impedance is 100 / (100 * 100) = 0.01 ohms. Assuming the arc drop remained the same, then the approximate arc power would be 50 * 50 / 0.01 = 250,000 W. In reality the arc drop will be higher and the current lower, but there will still be vastly more energy in this arc than the high voltage one.

.

 

[G._S._Ohm]

And, while we're on the subject. . . an arc is not a solid, a liquid or a gas. It is 'plasma', a so-called fourth state of matter.
It may exhibit negative resistance, incrementally. That is, as you increase the voltage the current decreases.

Striking an arc takes more voltage than maintaining an arc.

If you can get your hands on a discarded neon sign transformer you can learn a lot about high voltage, 30 mA arcs.