Arc-Flash: Proper I for determining Ct of Upstream OCPD

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MyCleveland

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Cleveland, Ohio
Which value is correct current in determining upstream OCPD clearing time ?
Simple 2 point system
Panel A has an OCPD that serves panel B.

bf = boltedfault af = arcing fault
Panel A : Ibf= 44,940.01 amps
Feeder to Panel B>>> 100’ - 1 set - #3's - Single conductors - Steel conduit

Panel B : Ibf= 10,221.30 amps, calc >>>> Iaf = 6,681.32 amps

REFLECT Iaf = 6,681.32 amps back to Panel A , calc>>>> Isc = 13,497.47 amps

Utilize Isc = 13,497.47 amps to derive CLEARING TIMEat OCPD at Panel A.
Agree or disagree ??
 
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wbdvt

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Rutland, VT, USA
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Electrical Engineer, PE
Per IEEE 1584, you also have to look at the trip time at 85% of arcing current to determine worst case for incident energy.
 
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smoothops10

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FL
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EE
Your Ibf to Iaf about Panel B looks good to me assuming IEEE 1584 method for arc in box at 0.48kV MCC/Panel w/ conductor gap of .25mm.

I don't understand what you are doing with the reflect calculation.

If Panel B has a fault with 6.7kA of arcing current that is the same current passing through the upstream OCPD at Panel A. Why use 13.5kA and how was that derived?
 
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Ingenieur

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smoothops10 said:
Your Ibf to Iaf about Panel B looks good to me assuming IEEE 1584 method for arc in box at 0.48kV MCC/Panel w/ conductor gap of .25mm.

I don't understand what you are doing with the reflect calculation.

If Panel B has a fault with 6.7kA of arcing current that is the same current passing through the upstream OCPD at Panel A. Why use 13.5kA and how was that derived?
Click to expand...

since the line is shorted a reflected wave of equal I magnitude is reflected, only slightly attenuated/dampened by the feeder Z
basically 2 x ifault
 
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smoothops10

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Ingenieur said:
since the line is shorted a reflected wave of equal I magnitude is reflected, only slightly attenuated/dampened by the feeder Z
basically 2 x ifault
Click to expand...

Not familiar with this. If I were to run this in SKM with 1584 method it would use the arcing fault current (af) - not 2 x af to calculate the trip time of upstream ocpd. I don't think it matters in OP's case as it is probably in the instantaneous region with both values for this scenario but obviosuly in a full study it would come up with quite different results since many ocpds would have quicker clearing times with the higher 2xaf being used.
 
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Ingenieur

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MyCleveland

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smoothops10 - you stated....
"If I were to run this in SKM with 1584 method it would use the arcing fault current (af) - not 2 x af to calculate the trip time of upstream ocpd."
Do you know this because SKM STATES the I value used in picking off Ct from TCC ? or are you assuming this is the case ?
 
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GoldDigger

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Ingenieur said:
this applies to transmission lines
theory for any conductor is similar
it's the only thing I can think of that relates
page 4 'short circuited line'
http://www.snscourseware.org/snsct/...Line Switching Overvoltages_November_2011.pdf

this is factored in HV T & D fault transient fault calcs, not so much LV 480 stuff
Click to expand...

:thumbsup:
If you do not have a transmission line of sufficient length (1 nanosecond per foot, roughly) the energy available from any possible short circuit impedance change will not be significant at the time scale of cycles of 60Hz.
 
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Ingenieur

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GoldDigger said:
:thumbsup:
If you do not have a transmission line of sufficient length (1 nanosecond per foot, roughly) the energy available from any possible short circuit impedance change will not be significant at the time scale of cycles of 60Hz.
Click to expand...

Hard to judge without ckt parameters
fault f = 1/(2Pi (LC)^2) could be several 100 Hz
surge Z = (L/C)^2
vel of prop = 1/(LC)^2, but usually 1/2 or so of c

but in the scale of bldg wiring moot
not familiar with arc flash calc, may be required?
 
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jim dungar

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MyCleveland said:
Where are we on the original question ?
Click to expand...

You do not need to consider any 'reflected values'. The person that first mentioned this condition went on to say they were not familiar with arc flash calculations.

Determine your bolted fault current. Use an industry accept methodology, like IEEE1584, to determine your arcing fault current. Use the arcing fault current to determine your operating time. Use the arcing current and opening time to determine your incident energy. Multiple calculations are likely required in order to cover device and system tolerances.
 
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