megloff11x
Senior Member
How is the bolted short circuit fault current determined for arc fault calculations?
arc flash - bolted short circuit current
- Thread starter megloff11x
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ryan_618
Senior Member
- Location
- Salt Lake City, Utah
Re: arc flash - bolted short circuit current
kVA/E/Z%
kVA/E/Z%
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
Re: arc flash - bolted short circuit current
I disagree with Ryan
[ February 01, 2006, 09:10 PM: Message edited by: ron ]
I disagree with Ryan
[ February 01, 2006, 09:10 PM: Message edited by: ron ]
- Location
- Illinois
- Occupation
- retired electrician
Re: arc flash - bolted short circuit current
This is the biggest problem with arc flash calculations...you can't get the information that is required to make the calculations, and as Ron said, a higher available fault current often results in a lower incident energy and a lower level of PPE.
Don
This is the biggest problem with arc flash calculations...you can't get the information that is required to make the calculations, and as Ron said, a higher available fault current often results in a lower incident energy and a lower level of PPE.
Don
Re: arc flash - bolted short circuit current
I work for a utility and can get instatanious information however this doesn't help because it constantly changes. Your utility will give you the maximum avalible fault current. However this is also a future number for things like transformer upgrades ect. What you really want is a realistic value for avalible fault current. I don't have an easy answer on how to get this.
charlie
Senior Member
- Location
- Indianapolis
Re: arc flash - bolted short circuit current
Neither do I.
Neither do I.
Re: arc flash - bolted short circuit current
Ryans method should not be used. You are looking for the time that the OC device will extinguish
the arc. This method give a high value that may lead to a false sense of security. Using the utility fault current may be the best of the worst case options. Therefore never take anything Ryan says as accurate.
Ryans method should not be used. You are looking for the time that the OC device will extinguish
the arc. This method give a high value that may lead to a false sense of security. Using the utility fault current may be the best of the worst case options. Therefore never take anything Ryan says as accurate.
ryan_618
Senior Member
- Location
- Salt Lake City, Utah
Re: arc flash - bolted short circuit current
Originally posted by bob:
Therefore never take anything Ryan says as accurate.
Re: arc flash - bolted short circuit current
I would like to understand this subject a little better.... I can't get a sensible answer from this equation: KVA/E/Z%
Assume 1000 kVA transformer @480 V
1000/480/5 = 0.416A
I don't mean to be petty, I imagine it's just a typo., but shouldn't it be VA instead of kVA? and the percentage expressed as a decimal? .05
1000000 / 480 /.05 = 41660
It also seems likely that it would be a 3 phase transformer (or bank) so the equation would become VA/(E*1.732)=A
1000000 / (480*1.732) = 1203A
With the 5% impedance rating, 1203A should flow in a shorted secondary with a voltage of 24(480 * 0.05) volts, so the impedance of the secondary:
24/1203=.01996 ohms
480/0.01996 = 24048 A
That's a little more than half of the previous equation. (probably about 1/1.732 )
Are you saying that the utility company's capacity to supply enough voltage or current to the primary to produce the full kVA rating of the transformer is unknown, and constantly changing; and therefore an arc could last longer, and there could be more heat produced?
I would like to understand this subject a little better.... I can't get a sensible answer from this equation: KVA/E/Z%
Assume 1000 kVA transformer @480 V
1000/480/5 = 0.416A
I don't mean to be petty, I imagine it's just a typo., but shouldn't it be VA instead of kVA? and the percentage expressed as a decimal? .05
1000000 / 480 /.05 = 41660
It also seems likely that it would be a 3 phase transformer (or bank) so the equation would become VA/(E*1.732)=A
1000000 / (480*1.732) = 1203A
With the 5% impedance rating, 1203A should flow in a shorted secondary with a voltage of 24(480 * 0.05) volts, so the impedance of the secondary:
24/1203=.01996 ohms
480/0.01996 = 24048 A
That's a little more than half of the previous equation. (probably about 1/1.732
)Are you saying that the utility company's capacity to supply enough voltage or current to the primary to produce the full kVA rating of the transformer is unknown, and constantly changing; and therefore an arc could last longer, and there could be more heat produced?
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
Re: arc flash - bolted short circuit current
You can simplify the calculation if you know the full load current at the secondary voltage. Take the full load current divided by the decimal form of the impedance (instead of 5% enter .05).
ie 1000kVA xfmr has 1204/.05 = 24kA of bolted fault current with an infinite amount of primary current available from the utility.
Since the utility has less than infinity amps on the primary for a fault event, the secondary current will be less than 24kA. Also consider impedances from cables and additional contribution from rotating masses within the facility that can be contributing in parallel.
You can simplify the calculation if you know the full load current at the secondary voltage. Take the full load current divided by the decimal form of the impedance (instead of 5% enter .05).
ie 1000kVA xfmr has 1204/.05 = 24kA of bolted fault current with an infinite amount of primary current available from the utility.
Since the utility has less than infinity amps on the primary for a fault event, the secondary current will be less than 24kA. Also consider impedances from cables and additional contribution from rotating masses within the facility that can be contributing in parallel.
- Location
- Illinois
- Occupation
- retired electrician
Re: arc flash - bolted short circuit current
Yes, the calculation shown in this thread gives the short circuit current at the secondary of the transformer based on an infinite utility primary. The fault current will always be some what less than this calculated value.
Re: arc flash - bolted short circuit current
It seems to me a lot of times there is no definite answer, but a sort of range finding. What you read on your volt meter is seldom nominal voltage. Everything's kinda dynamic.
Would there be a likely "range" of primary current available, rather than infinite? Could you "reconcile" the overcurrent device's clearing curve against that value? Or is that what some of you are proposing rather than using the simple equation?
If you're talking about the transformer's bolted fault current, aren't you also talking about everything else connected down line?
[ February 05, 2006, 07:15 AM: Message edited by: realolman ]
It seems to me a lot of times there is no definite answer, but a sort of range finding. What you read on your volt meter is seldom nominal voltage. Everything's kinda dynamic.
Would there be a likely "range" of primary current available, rather than infinite? Could you "reconcile" the overcurrent device's clearing curve against that value? Or is that what some of you are proposing rather than using the simple equation?
If you're talking about the transformer's bolted fault current, aren't you also talking about everything else connected down line?
[ February 05, 2006, 07:15 AM: Message edited by: realolman ]
charlie
Senior Member
- Location
- Indianapolis
Re: arc flash - bolted short circuit current
This is the statement given on GB7-060 in our table of Standard Calculated Fault Currents, "The standard calculated available fault currents are given in amperes, RMS symmetrical, at the secondary bushings of the Company's transformer, assuming an infinite bus and a bolted fault. The intent of these values is to serve as a guide in the selection of the proper service and downstream equipment. These are of no value for the use in determining the proper personal protection equipment since the impedance, fuse size and type, and calculated primary fault current available at the primary bushings of the bank can not be known for any particular installation."
The formula that Ryan gave you is how this table was derived and the disclaimer is because the incident energy is also a function of time. Assuming the amount of fault current will be less that we tell you (it will be), the current limiting device will operate slower and could actually permit a higher incident energy to be available.
This is the statement given on GB7-060 in our table of Standard Calculated Fault Currents, "The standard calculated available fault currents are given in amperes, RMS symmetrical, at the secondary bushings of the Company's transformer, assuming an infinite bus and a bolted fault. The intent of these values is to serve as a guide in the selection of the proper service and downstream equipment. These are of no value for the use in determining the proper personal protection equipment since the impedance, fuse size and type, and calculated primary fault current available at the primary bushings of the bank can not be known for any particular installation."
The formula that Ryan gave you is how this table was derived and the disclaimer is because the incident energy is also a function of time. Assuming the amount of fault current will be less that we tell you (it will be), the current limiting device will operate slower and could actually permit a higher incident energy to be available.
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