1000A SERVICE FEED

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Toros

Senior Member
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Tujunga, CA
HI, IS,
(12) 3" EMT, (4) 400 MCM CU IN EACH IS GOOD SIZE
TO FEED A NEW 1000A, 277/480V, 4-WIRE DISTRIBUTION BOARD
(FEEDER ON THE ROOF, WHERE TEMP. DE-RATING APPLIES"???

TANX
 
You have well over 1000 amps of conductor before adjustments, but gave us no detail whatsoever to let us know what adjustments need made other then it is on a roof.
 
(12) 400 kcmilCu will be good for 1000 amps regardless of the conditions. Go for it !
 
infinity said:
Sounds more like a mistake than an actual good design. Of course there could be other reasons for such overkill. :)
Click to expand...
Future expansion? But one could run raceway now and pull conductors when needed later on, and save just a couple bucks.
 
Toros said:
HI, IS,
(12) 3" EMT, (4) 400 MCM CU IN EACH IS GOOD SIZE
TO FEED A NEW 1000A, 277/480V, 4-WIRE DISTRIBUTION BOARD
(FEEDER ON THE ROOF, WHERE TEMP. DE-RATING APPLIES"???

TANX
Click to expand...


I MADE BAD DESIGN, SORRY
I NEED THIS,

(4) 3" EMT , (4) # 300 MCM + #4 EQ. GR. THHN, CU IN EACH COND.
 
Toros said:
I MADE BAD DESIGN, SORRY
I NEED THIS,

(4) 3" EMT , (4) # 300 MCM + #4 EQ. GR. THHN, CU IN EACH COND.
Click to expand...

That sounds better. ;)

So you have 4 sets of 300's @ 75? C = 4*285 amps = 1140 amps which is good for 1000 amp load. For your roof top correction you can use the 90? C ampacity of 4*320 amps = 1280 amps and work down from there.

Also your #4 EGC too small for a 1000 amp load, from table 250.122 the minimum size required in each parallel raceway is #2/0.
 
infinity said:
That sounds better. ;)

So you have 4 sets of 300's @ 75? C = 4*285 amps = 1140 amps which is good for 1000 amp load. For your roof top correction you can use the 90? C ampacity of 4*320 amps = 1280 amps and work down from there.

Also your #4 EGC too small for a 1000 amp load, from table 250.122 the minimum size required in each parallel raceway is #2/0.
Click to expand...
TANX
I FEEL BETTER
 
infinity said:
That sounds better. ;)

So you have 4 sets of 300's @ 75? C = 4*285 amps = 1140 amps which is good for 1000 amp load. For your roof top correction you can use the 90? C ampacity of 4*320 amps = 1280 amps and work down from there.

Also your #4 EGC too small for a 1000 amp load, from table 250.122 the minimum size required in each parallel raceway is #2/0.
Click to expand...

I agree with this up til recently where we now consider the neutral a CCC (not considering the roof derating). The safer bet is to assume the neutral is a CCC. Therefore, 1000A/.8 = 1,250A to design to based on the 90-degree column. 4 sets of 300's @ 90? C = 4*320 amps = 1280 amps which is higher than the derated amp draw so the 300's are still good. Or, use the 90-deee column... 320A*.8*4 =1024A>1000A so good to go. This is per 310.15.B.5.C and almost everything nowadays is non-linear. I think the best bet anymore is derate for 4CCC then there is no debate. However, this may start a debate. Now you start drating for the roof and you may need to go up a size.

Also, if this is off of a step-down transformer for some reason, the ground needs to be based on table 250.66 which would require a 3/0 ground. This is per 250.30.A.2 which refers to 250.102.C which refers to table 250.66. Exhibits 250.13 and 250.14 shows you the ground from the transformer is a bonding jumper.
 
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mjmike said:
I agree with this up til recently where we now consider the neutral a CCC (not considering the roof derating). The safer bet is to assume the neutral is a CCC. Therefore, 1000A/.8 = 1,250A to design to based on the 90-degree column. 4 sets of 300's @ 90? C = 4*320 amps = 1280 amps which is higher than the derated amp draw so the 300's are still good. Or, use the 90-deee column... 320A*.8*4 =1024A>1000A so good to go. This is per 310.15.B.5.C and almost everything nowadays is non-linear. I think the best bet anymore is derate for 4CCC then there is no debate. However, this may start a debate. Now you start drating for the roof and you may need to go up a size.

Also, if this is off of a step-down transformer for some reason, the ground needs to be based on table 250.66 which would require a 3/0 ground. This is per 250.30.A.2 which refers to 250.102.C which refers to table 250.66. Exhibits 250.13 and 250.14 shows you the ground from the transformer is a bonding jumper.
Click to expand...


What do you base this on? IMO unless this switchboard is dedicated to a majority of specific types of non-linear loads there is no reason to count the neutral as a CCC in the feeder.
 
infinity said:
What do you base this on? IMO unless this switchboard is dedicated to a majority of specific types of non-linear loads there is no reason to count the neutral as a CCC in the feeder.
Click to expand...

I am not basing it on anything contained in this thread. I am basing it on what is mostly seen now. If you have a motor, it most likely has a VFD. IF you have lights, they are most like non-linear. Basically, the only things I can think of that wouldn't cause this to be non-linear would be incandescent lights and resistive heating elements. The code is somewhat ambiguous by stating "a majority" so it is up for grabs, but it might be safer to design for that now. Just offering my opinion. Like I said, I agree with everything posted until the debate of the neutral comes into play. Maybe we need the OP to tell us if the loads are linear or non-linear?
 
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Some loads that used to be very non linear, like VFD inputs and various lighting with electronic ballasts or drivers, are now available with high PF input stages. This can mitigate the triplen harmonics to the point where the neutral current for a balanced load remains small.
 
A typical three phase VFD is a non linear load, but it is not drawing any neutral current either. If it is supplied from a wye connected source the current does pass through the neutral point of that source.
 
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