Transformer available fault current

jim dungar

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Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Can you briefly mention the math behind that?
High fault currents may cause the protective device to operate as fast as possible so "t" is very small, even into sub-cycle time. Low fault current may cause the protective device in its short time region so "t" might be measured in seconds.

My experience has been the majority locations with AFIE over 25 cal/cm*2 is due to long fault clearing times and not high available fault currents.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
High fault currents may cause the protective device to operate as fast as possible so "t" is very small, even into sub-cycle time. Low fault current may cause the protective device in its short time region so "t" might be measured in seconds.
OK, that's for large difference in fault current causing a change between regions of the trip curve.

For a small change in fault current, so that the region of the trip curve is unchanged, is more fault current always more energy? Or can you have a region of the trip curve where locally the energy released varies inversely with fault current?

Thanks,
Wayne
 

jim dungar

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Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
OK, that's for large difference in fault current causing a change between regions of the trip curve.

For a small change in fault current, so that the region of the trip curve is unchanged, is more fault current always more energy? Or can you have a region of the trip curve where locally the energy released varies inversely with fault current?

Thanks,
Wayne
I have seen changes of less than 1000A move from >40 cals to less than 1.2 cals. It all depends on which side of the instanteous opening curve you fall on.
 

steez

Member
Location
TX
Occupation
Electrical PE
Can you briefly mention the math behind that? A statement like (made up, probably wrong) "in some regions of possible fault current, the opening time will vary as 1/I^2, while the power delivered during that time will vary as I, so the energy released varies as 1/I."

Thanks,
Wayne


Hi Wayne,
Sorry for the delay, I am happy to try and explain. For the equations, always refer to NFPA 70E and IEEE. They have changed slightly over the years (for example, they now have electrode configuration of the panel bussing)

I think you already understand some of this, given you indicated in a previous comment that the magnitude of the difference in current must cause the breaker to operate in a different portion of the trip curve. That's it right there. If the time to trip is held constant, the higher the magnitude of the current, the higher the IE.
 
Last edited:

David Castor

Senior Member
Location
Washington, USA
Occupation
Electrical Engineer
Wayne,

Arc-flash incident energy is mainly a function of the magnitude of the arcing current and how long it lasts (time). If you think about a molded case breaker, it has two protective functions - the long-time (thermal) and the "instantaneous" or magnetic. If the arc current is such that this breaker will trip on the thermal curve, the arc-time could be two seconds (generally used as maximum arc-time) or much longer in theory. If the fault current is higher and the breaker trips on the magnetic trip, the arc-time might be 3 cycles or less.

For arc-flash in the 480 V world, if you are downstream of a molded-case breaker tripping on the magnetic curve, the incident energy will be manageable almost regardless of the available fault current. It's really difficult to get over 8 cal/cm2. But if the upstream breaker is tripping on its thermal curve, you're screwed because it is so much slower. This is why incident energy when running on a standby generator is often WORSE than when connected to the utility.

I can send you a couple of examples if you're interested.

Cheers,

Dave
 
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