Service conductors

[hhsting]

I have service conductors from POCO transformer run underground inside the building utility room that has 4000A service disco.

The demand load at 4000a service disconnect is 3200A.

I have email from POCO representative stating that POCO would bring in from POCO transformer 7 sets of 500kcmil AL conductors to the line side terminal of 4000A service disconnect and terminate the conductors at line side terminal of the service disco.

Now I need to size the main bonding jumper and bonding jumper to enclosure and their size depend on the incoming service conductor.


Questions:

1. Would this make the service point at the 4000A service disconnect or not?

2. Are not the service conductors 7 sets of 500kcmil AL small when compared with demand load of 3200A at the 4000A service disco or NEC would not apply sizing those conductors?

 

[Hv&Lv]

I have service conductors from POCO transformer run underground inside the building utility room that has 4000A service disco.

The demand load at 4000a service disconnect is 3200A.

I have email from POCO representative stating that POCO would bring in from POCO transformer 7 sets of 500kcmil AL conductors to the line side terminal of 4000A service disconnect and terminate the conductors at line side terminal of the service disco.

Now I need to size the main bonding jumper and bonding jumper to enclosure and their size depend on the incoming service conductor.


Questions:

1. Would this make the service point at the 4000A service disconnect or not?

2. Are not the service conductors 7 sets of 500kcmil AL small when compared with demand load of 3200A at the 4000A service disco or NEC would not apply sizing those conductors?

1. If that is where the utility has their delivery point, then yes.
2. No. Not of this is the point of delivery.

 

[Dennis Alwon]

The size of the conductors from the power company is not your worry but it is very small.
I would base my service bonding jumper on the full 4000 amps using maybe 9 sets of 1000kcm.

 

[hhsting]

The size of the conductors from the power company is not your worry but it is very small.
I would base my service bonding jumper on the full 4000 amps using maybe 9 sets of 1000kcm.

I thought Table 250.102(C)(1) one needs to size main bonding jumper based on largest ungrounded conductor or equivalent area for parallel conductors which would be in my case 7 sets of 500kcmil Al provided which is small but they are from POCO

I didnt know to size main bonding jumper based on service size 4000A and not largest ungrounded conductor. So why not other conductors besides 9 sets of 1000kcmil? I mean other combination can also be used how did you arrive at 9 sets of 1000 kcmil?

 

[Dennis Alwon]

I just chose 9- 1000 kcm's for 4000 amps. I don't think you can use the undersized cables of the power company so I would invoke note 3 and 4 of table 250.102(C)(1)

 

[Little Bill]

I thought Table 250.102(C)(1) one needs to size main bonding jumper based on largest ungrounded conductor or equivalent area for parallel conductors which would be in my case 7 sets of 500kcmil Al provided which is small but they are from POCO

I didnt know to size main bonding jumper based on service size 4000A and not largest ungrounded conductor. So why not other conductors besides 9 sets of 1000kcmil? I mean other combination can also be used how did you arrive at 9 sets of 1000 kcmil?

Are you a designer or plans reviewer?
Seems to me if you are a reviewer someone should submit THEIR plans and YOU just review them.

 

[electrofelon]

I just chose 9- 1000 kcm's for 4000 amps. I don't think you can use the undersized cables of the power company so I would invoke note 3 and 4 of table 250.102(C)(1)

Dennis I agree note 4 would apply here, but why note 3?

 

[augie47]

If it's a factory service rated disconnect the manufacturer should have provided the main bonding jumper.
If not, to cover yourm rear, I would do as Dennis sugest and size it based on 9-1000s. If there is a problem later, let it be POCOPs and not yours.

 

[hhsting]

I just chose 9- 1000 kcm's for 4000 amps. I don't think you can use the undersized cables of the power company so I would invoke note 3 and 4 of table 250.102(C)(1)

Gotcha but one can also then base it on 13 sets of 500kcmil Al to size the main bonding jumper no? Main bonding jumper is wire

 

[Dennis Alwon]

Dennis I agree note 4 would apply here, but why note 3?

Note 3 just states to add the multiple sets of conductors together. Since there isn't any one conductor large enough for 4000 amps then I would use note 3 to add the cm together. In my example it would be 9000 kcm

 

[electrofelon]

Gotcha but one can also then base it on 13 sets of 500kcmil Al to size the main bonding jumper no? Main bonding jumper is wire

it is interesting because note 4 isnt really clear on how to calculate the size. Using 1/0, the minimum size allowed for parallels would give youthe most ampacity per cmil which would be 22 sets or a total of 2323 KCMIL so then 12.5% of that is 290 KCMIL. Using 750 would take a total of 7 sets or 5250 KCMIL and 12.5% of that is 656 KCMIL.

 

[electrofelon]

Note 3 just states to add the multiple sets of conductors together. Since there isn't any one conductor large enough for 4000 amps then I would use note 3 to add the cm together. In my example it would be 9000 kcm

But note 3 is talking sets specifically for 230.40 Ex #3 not parallels in general, I believe.

 

[hhsting]

But note 3 is talking sets specifically for 230.40 Ex #3 not parallels in general, I believe.

Note #3 Does it not say 230.40 exception no. 2?

 

[Dennis Alwon]

I think it applies. What would you do with 9 sets of 1000 kcm conductors. Wouldn't you add the 9 kcm's together to get the equivalent size? That is what #3 is saying, IMO

or if multiple sets of ungrounded supply
conductors are installed for a separately derived system

 

[Dennis Alwon]

Note #3 Does it not say 230.40 exception no. 2?

Yes but it also says OR--- that is my opinion-- I'm going to bed. Should have been there an hour ago

If multiple sets of service-entrance conductors are used as permitted
in 230.40, Exception No. 2, or if multiple sets of ungrounded supply
conductors are installed for a separately derived system, the equivalent
size of the largest ungrounded supply conductor(s) shall be
determined by the largest sum of the areas of the corresponding
conductors of each set.

 

[hhsting]

it is interesting because note 4 isnt really clear on how to calculate the size. Using 1/0, the minimum size allowed for parallels would give youthe most ampacity per cmil which would be 22 sets or a total of 2323 KCMIL so then 12.5% of that is 290 KCMIL. Using 750 would take a total of 7 sets or 5250 KCMIL and 12.5% of that is 656 KCMIL.

So Dennis just chose randomly 1000 kcmil? One can use any size cable?


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[electrofelon]

Yes but it also says OR--- that is my opinion-- I'm going to bed. Should have been there an hour ago

If multiple sets of service-entrance conductors are used as permitted
in 230.40, Exception No. 2, or if multiple sets of ungrounded supply
conductors are installed for a separately derived system, the equivalent
size of the largest ungrounded supply conductor(s) shall be
determined by the largest sum of the areas of the corresponding
conductors of each set.

I do not believe we are talking about an SDS dennis.

Yes HHSTing I meant #2 not #3

Either way its sorta a moot point, as augie said, this gear will have a factory MBJ

 

[electrofelon]

So Dennis just chose randomly 1000 kcmil? One can use any size cable?


Sent from my iPhone using Tapatalk

That is the problem, it doesnt seem to say how to select your sets and Kcmil does not have a linear relationship to ampacity

 

[hhsting]

I do not believe we are talking about an SDS dennis.

Yes HHSTing I meant #2 not #3

Either way its sorta a moot point, as augie said, this gear will have a factory MBJ

Its existing piece of gear they are doing utility upgrade due to load changes going up downstream. Existing gear has existing wire size MBJ

 

[hhsting]

That is the problem, it doesnt seem to say how to select your sets and Kcmil does not have a linear relationship to ampacity

Thats a big problem. One can use 2000 kcmil Al 8 sets and get MBJ of 2000 kcmil