800Amps Circuit Breaker with (2) 500Kcmil Lugs

[faresos]

Hello Everyone;

I'm working on a project where the contractor needs to provide a combination main circuit breaker with CT section (model # Erickson CTA-MD-467N-6CM8) per the utility standard. The line side was called out to terminate (2) sets of 500Kcmil, however, the load side was called out for (3) sets of 300Kcmil. The contractor saying that we have the wrong quantity of lugs on the load side of the breaker. my question, can the breakers lugs on the load side be change out from (2) sets of 500kcmil to (3) sets of 300kcmil?

Thanks,

 

[texie]

I don't see how you how you have a factory breaker with lugs that won't even allow 800 amps worth of conductor. 2 X 500 MCM = 760 amps. There must be more to this.

 

[Rock86]

Just to clarify, the load side of the CT is (2) sets of 500Kcmil to the line side of the breaker, and you want (3) sets of 300Kcmil out from the breaker? The CT you spec'd says it can handle up to (3) sets of 600 on the load side so you shouldn't have an issue there. What breaker was spec'd?

 

[don_resqcapt19]

Most breakers have interchangeable lugs. You should be able to buy a lug kit from the breaker manufacturer and field replace the existing lugs.

 

[faresos]

Just to clarify, the load side of the CT is (2) sets of 500Kcmil to the line side of the breaker, and you want (3) sets of 300Kcmil out from the breaker? The CT you spec'd says it can handle up to (3) sets of 600 on the load side so you shouldn't have an issue there. What breaker was spec'd?

I think you are looking at the fuse disconnect type not the circuit breaker type which indicated to be (2) sets of 500kcmil.

 

[roger]

As Don said, most large frame breakers have interchangeable lugs.

Roger

 

[infinity]

I think you are looking at the fuse disconnect type not the circuit breaker type which indicated to be (2) sets of 500kcmil.

According to the model number in the OP the CB type does say that it accepts 2-500 kcmil.

http://products.ericksonelectric.com/Asset/CT%20&%20DISCONNECT%20(CT%20BOTTOM)%20-%206CM-CTB.pdf

 

[texie]

According to the model number in the OP the CB type does say that it accepts 2-500 kcmil.

http://products.ericksonelectric.com/Asset/CT%20&%20DISCONNECT%20(CT%20BOTTOM)%20-%206CM-CTB.pdf

I don't understand why they list lugs that don't have enough ampacity. What am I not getting here?

 

[don_resqcapt19]

I don't understand why they list lugs that don't have enough ampacity. What am I not getting here?

Because parallel 500 kcmil copper conductors is a very common feed for an 800 amp circuit and is permitted by 240.4(B) as long as the calculated load is 760 amps or less.

Since these appear to be service conductors, 230.90(A), Exception #2 specifically permits the use of the rule in 240.4(B).

 

[texie]

Because parallel 500 kcmil copper conductors is a very common feed for an 800 amp circuit and is permitted by 240.4(B) as long as the calculated load is 760 amps or less.

Since these appear to be service conductors, 230.90(A), Exception #2 specifically permits the use of the rule in 240.4(B).

Well, yes, all true but it just seems odd that they would force you to change lugs if your calculated load is above 760. After all, it is an 800 amp breaker and you would think that they would give you at least 2 600 lugs.

 

[don_resqcapt19]

Well, yes, all true but it just seems odd that they would force you to change lugs if your calculated load is above 760. After all, it is an 800 amp breaker and you would think that they would give you at least 2 600 lugs.

Typically, the lugs are specified when you order the equipment. It is on the person placing the order. Not sure what the default lug sizing would be, as most manufacturers do have a default that they provide when a specific lug is not specified. For this application, I would expect that the default for a two conductor lug would be 600 kcmil, but not sure.

 

[petersonra]

It is up to the people specifying the equipment to decide which of the available lugs should be used. Some breakers come with no lugs at all and they have to be ordered. Others come with a default set of lugs that can be removed and something more suitable used. Sometimes the default part number for a breaker comes with lugs and sometimes it doesn't. you have to do a little bit of thinking and research when you order them.

For instance, one series of Siemens breakers comes with lugs. Another series you have to add an l to the part number to get the default lugs. Or you can just order the lug separately and throw away the ones they come with because as often as not, it is faster to get the default part number and sometimes cost less then getting the breaker that has no lug at all.

 

[paulengr]

Most breakers of this size these days come STANDARD with 4 terminals per phase rated for up to 350 kcmil.

Plus it probably doesn’t matter. 800 is the frame size and will have an adjustable trip. Since it’s distribution you should use a UL 1077 breaker rather than UL 489 so you can coordinate with downstream.

There is an obvious reason why the breaker comes standard this way. One 350 (more common than 300 which is sort of oddball) is rated 310 A. So without derating we are already well over 900 A at 75 C, where we can’t even get to 800 A at dual 500s. A 500 is rated 380 A off the 75 C chart. So those two 500s get you to 760. In a single conduit we get 344 each so we are at only 692 A. On the 350s we can pack all 9 in there and be at 245 A each or 735 A total, more than the 6 500s or split things into separate conduits. And the conduit size (fill) is nearly the same. Either way the implication is very clear...350 gives you as much or more ampacity as 500s. But wait there’s more. Using wirecableyourway.com just for quick comparisons 350 is currently $5.91/foot. 500 is $8.62/foot. So nine 360s runs you $53.19/foot and 6 500s runs you $51.42/foot, so roughly even cost. However the labor factor is much higher with the 500s since it often requires special tools and a lot more work installing it. We figure 25-50% more on pulls. As in nearly anyone can bend and route 350 but with 500 often you have to break out the cable benders. And if triple 300s is sufficient it’s only $5.09 or $45.36/foot. If you stick with 3/C per raceway granted the extra raceway cost eats into this a little. But if you run one big raceway or use tray, it’s a nonissue. With multiple raceways you still can’t get there with 500 and you only need one extra conduit with 350. So even if you are running truly at 800 A of load, the triple 350 will give you more than two 500s will. Hence the breakers are set up that way for that reason, because this is the most common configuration in that size.

If you truly can’t switch lugs this is easily fixed. Buy power distribution blocks. They’re cheap. Get 2 in and 4 out. Or if you have resources to do it use busbars. We do it constantly with motors because there is almost always a motor lead/feeder mismatch. Motors are typically leaded up to 4/0 maximum.

 

[infinity]

I don't see how you how you have a factory breaker with lugs that won't even allow 800 amps worth of conductor. 2 X 500 MCM = 760 amps.

I agree, the CB should have a minimum standard lug size of (2)-600 kcmil for 800 amps.

 

[kwired]

800 amps is a point where one may choose two or three sets of conductors for paralleling, I would expect there to be lug options on nearly all 800 amp and higher devices.

 

[Carultch]

Most breakers of this size these days come STANDARD with 4 terminals per phase rated for up to 350 kcmil.
......

Understood that there is a principle advantage of using more parallel sets, due to more amps per kcmil of conductor, and wires that are significantly easier to work with mechanically. It has to do with the perimeter to area ratio of the circles when compared, because the greater that ratio, the more surface area there is for the heat to escape. That first advantage only really applies, when length isn't significant enough to make voltage drop matter, because when voltage drop matters, you generally need about the same kcmil to reduce the total ohms of the circuit, regardless of how the kcmil of conductor is distributed. To a lesser extent, the inductive effects of effective AC resistance (ch 9/ Table 9) in wires give you a slight advantage, but the advantage is a lot less significant than the factors that govern the minimum local size.

Still, this is missing the OP's point about the sizes given for this particular breaker. You would think that a breaker would be built with terminal capacity to accept conductors that at minimum can meet its own trip rating. The OP's example doesn't allow any conductor combination that adds up to the full 800A. Only 2x500's which can at most add up to 760A and require you to take credit for 240.4(B). Or similarly, 310.15(B)(7), in cases where that rule applies. What if you were using this breaker as the destination for tap conductors, where it is mandatory to get the full 800A? What if you calculated load exceeds 760A? You couldn't use this breaker.

As other's have mentioned, it is probably a result of the original installer customizing the breaker lug selection, in an application where you can take credit for 240.4(B).

 

[kwired]

Understood that there is a principle advantage of using more parallel sets, due to more amps per kcmil of conductor, and wires that are significantly easier to work with mechanically. It has to do with the perimeter to area ratio of the circles when compared, because the greater that ratio, the more surface area there is for the heat to escape. That first advantage only really applies, when length isn't significant enough to make voltage drop matter, because when voltage drop matters, you generally need about the same kcmil to reduce the total ohms of the circuit, regardless of how the kcmil of conductor is distributed. To a lesser extent, the inductive effects of effective AC resistance (ch 9/ Table 9) in wires give you a slight advantage, but the advantage is a lot less significant than the factors that govern the minimum local size.

Still, this is missing the OP's point about the sizes given for this particular breaker. You would think that a breaker would be built with terminal capacity to accept conductors that at minimum can meet its own trip rating. The OP's example doesn't allow any conductor combination that adds up to the full 800A. Only 2x500's which can at most add up to 760A and require you to take credit for 240.4(B). Or similarly, 310.15(B)(7), in cases where that rule applies. What if you were using this breaker as the destination for tap conductors, where it is mandatory to get the full 800A? What if you calculated load exceeds 760A? You couldn't use this breaker.

As other's have mentioned, it is probably a result of the original installer customizing the breaker lug selection, in an application where you can take credit for 240.4(B).

Plus you would need even larger conductors if using aluminum.

 

[faresos]

Thank you all for your responses. One more question, if for any reason the contractor came back saying this is cannot be done (changing the lugs), is there a connector (splice kit) can be used to terminate incoming (3) 300kcmil to (2) 500kcmil before terminating into the breaker lugs? Not sure if there is adequate space in that main disconnect enclosure to do this but I guess a splice box can be used.

 

[petersonra]

How would that help you if two 500 s do not give you enough ampacity? Polaris and Panduit makes such connectors but I don't think they're going to solve the problem.

 

[electrofelon]

I did not reread the whole thread, but I don't think it has been established that OP necessarily needs the full 800 A, rather he has this conductor combination that someone has spec'd. If he does need 800A ampacity, What about splicing the three conductors to two 600's and using pin reducers? Obviously try reluging the breaker first.