Arc Flash Calc

Isaiah

Senior Member
I have a 480V, 3Ph, 3W switchrack located in a CID2 area - Refinery. SWRK has a 200AT/250AF MCB along with 2-40Hp motors, 2-7.5Hp motors and a 30kVA transformer/panelboard feeding mainly HID lighting loads. Client wants to add two more 460V motors at 15Hp each. Is it necessary to revisit the Arc Flash study and modify it due to these two additional loads?
 

wsbeih

Member
I have a 480V, 3Ph, 3W switchrack located in a CID2 area - Refinery. SWRK has a 200AT/250AF MCB along with 2-40Hp motors, 2-7.5Hp motors and a 30kVA transformer/panelboard feeding mainly HID lighting loads. Client wants to add two more 460V motors at 15Hp each. Is it necessary to revisit the Arc Flash study and modify it due to these two additional loads?
Per NFPA 70E, 130.5 (G): "The incident energy analysis shall be updated when changes occur in the electrical distribution system that could affect the results of the analysis. The incident energy analysis shall also be reviewed for accuracy at intervals not to exceed 5 years."
 

wbdvt

Senior Member
Per NFPA 70E, 130.5 (G): "The incident energy analysis shall be updated when changes occur in the electrical distribution system that could affect the results of the analysis. The incident energy analysis shall also be reviewed for accuracy at intervals not to exceed 5 years."
True but IEEE 1584 only considers motors larger than 50 hp. But let's look at this from a practical point.

A 460V 15hp motor will contribute about 70A of short circuit current at the motor terminals so fault current will less at MCC due to conductor impedance and last only about 1/2 to 1 cycle. So if the available bolted fault current is 15kA at the MCC, the fault current from the motor is <0.5%. This will not make a difference in the incident energy at the MCC as the arcing current clearing time is greater than 1/2 cycle and the source current is the major determiner of the incident energy.
 

iceworm

Curmudgeon still using printed IEEE Color Books
Following wb's thoughts:
Working under cumulative motor fault current,
2 x 15hp = 30 hp >> FLA = 40A
Induction motor fault contribution of 4 x FLA = 160A, Close to WB's estimate.

As long as we are not getting just half the story.

Panel loading:
1 ea 40 hp (at 125%) = 65A
1 ea 40 hp (100%) = 52A
2 ea 7.5 hp = 22A
30kva at 80% = 29A​

Panel load before addition of 15hp motors = 168A
And if this is continuous load then 168 x 1.25 = 210A

Adding:
2 ea 15 hp = 42A​

Panel load after addition = 210A
If continuous load = 210 x 1.25 = 262A

And this is all on a panel with a 200A trip Main CB.

What am I missing?
 

wbdvt

Senior Member
Following wb's thoughts:
Working under cumulative motor fault current,
2 x 15hp = 30 hp >> FLA = 40A
Induction motor fault contribution of 4 x FLA = 160A, Close to WB's estimate.

As long as we are not getting just half the story.

Panel loading:
1 ea 40 hp (at 125%) = 65A
1 ea 40 hp (100%) = 52A
2 ea 7.5 hp = 22A
30kva at 80% = 29A​

Panel load before addition of 15hp motors = 168A
And if this is continuous load then 168 x 1.25 = 210A

Adding:
2 ea 15 hp = 42A​

Panel load after addition = 210A
If continuous load = 210 x 1.25 = 262A

And this is all on a panel with a 200A trip Main CB.

What am I missing?
Well, probably not all the info to determine if loading violates NEC because the OP specifically asked about the need to redo the arc flash study and that is how I answered the question. Granted there may be other considerations to be taken into account but none of those were asked.
 

Isaiah

Senior Member
Well, probably not all the info to determine if loading violates NEC because the OP specifically asked about the need to redo the arc flash study and that is how I answered the question. Granted there may be other considerations to be taken into account but none of those were asked.

The motor loads are connected to pumps that are intermittent duty; plus each pair of motors is a primary/backup scenario, so only one pump operates at one time. The transformer serves mainly lighting loads so this load is continuous.
 

iceworm

Curmudgeon still using printed IEEE Color Books
That was just curiosity on my part. It didn't add up.

So, as for the arc-flash, you are adding only one 15hp motor. The other is another primary/backup? If so, that only adds half, ~80A to the SCC.
I don't see any reason to redo the arc-flash
 

Isaiah

Senior Member
That was just curiosity on my part. It didn't add up.

So, as for the arc-flash, you are adding only one 15hp motor. The other is another primary/backup? If so, that only adds half, ~80A to the SCC.
I don't see any reason to redo the arc-flash
Yes, the two additional motors are also primary and backup pumps. I greatly appreciate your input.
Isaiah
 

MyCleveland

Senior Member
You need to add a LABEL to each item added indicating SCC.
Assuming you have the software for that, only one more step to run the Arc-Flash numbers.
 

Isaiah

Senior Member
Following wb's thoughts:
Working under cumulative motor fault current,
2 x 15hp = 30 hp >> FLA = 40A
Induction motor fault contribution of 4 x FLA = 160A, Close to WB's estimate.

As long as we are not getting just half the story.

Panel loading:
1 ea 40 hp (at 125%) = 65A
1 ea 40 hp (100%) = 52A
2 ea 7.5 hp = 22A
30kva at 80% = 29A​

Panel load before addition of 15hp motors = 168A
And if this is continuous load then 168 x 1.25 = 210A

Adding:
2 ea 15 hp = 42A​

Panel load after addition = 210A
If continuous load = 210 x 1.25 = 262A

And this is all on a panel with a 200A trip Main CB.

What am I missing?
I think you've factored the continuous loads too many times by 125%. You only need to do this once - with the largest motor - and once again with non-motor loads, i.e. transformer (lighting loads).
 

wbdvt

Senior Member
You need to add a LABEL to each item added indicating SCC.
Assuming you have the software for that, only one more step to run the Arc-Flash numbers.
Why? Some things don't get labels like motor terminal boxes. The short circuit current only needs to be on the service entrance equipment.
 

iceworm

Curmudgeon still using printed IEEE Color Books
I think you've factored the continuous loads too many times by 125%. You only need to do this once - with the largest motor - and once again with non-motor loads, i.e. transformer (lighting loads).
You're right. Yuck - sloppy me.
 

wbdvt

Senior Member
Plus no state has adopted the 2020 NEC yet. And the definition of industrial machinery does not seem to be applicable to a stand alone motor.
 
Top