Paralleled Installation

[Smart $]

Yes, my assumptions may be more synical, but we both are guilty of assuming.

Yes, it is an assumption, for EMT can be used underground... I'm just weighting the assumption with likelihood.

Yes, my grammer is bad, but my math is the same as last year. We looked up IEEE formulas on temp rise a few years back. You liked my results then, and I'm using the same formulas on my spreadsheet now. The label on my spreadsheet actual reads "+Temp Rise." My ambient is adjustable, currently set at 30c.

The question I have is, are you stating T2 or ΔT.

Temperature Rise is ΔT only.

T2 = T0 + ΔT​

...where T0 is the ambient temperature. (Not shown is T1, which is 75?C, the conductor temperature for values given in Table 8 and 9.)

...If my calcs are wrong please help me fix it.

We first have to establish if it is just an error in how you are stating it. Otherwise, it would be easiest to determine any error by seeing your calc's.

 

[don_resqcapt19]

Correction: 560A could burn up four 350's. I forgot to derate, "four" CCC's. Operating a non-continuous 560A load is a 90c temp. rise (when four CCC's share same conduit)., and 4.6% voltage drop over 500 feet.

Roger,
If 560 amps can burn up parallel 350s with 4 current carrying conductors in a raceway, then Table 310.16 and the associated adjustment factors have some big problems. That is the permitted ampacity of 350 THHN copper under those conditions.

 

[ramsy]

My use of term "burn up" refers to equipment listings. I assume most lugs, boxes, & enclosures are not listed for 90c, much less for DC rated for continuous use at 124c.

If we don't agree on 90c as the operating temp., then lets look at my derating first.
Two #350's = 2 x 350A @ 90c or (700A * 0.8) with four CCC's = 560A @ 90c.

So a full 560A load operates at 90c, if we count all DC conductors as CCC's. The positive polarity is two cables + the negative polarity is two cables, for a total of four CCC's.

 

[ramsy]

The question I have is, are you stating T2 or ΔT. ..We first have to establish if it is just an error in how you are stating it. Otherwise, it would be easiest to determine any error by seeing your calc's.

Please check my 90c operating results for this 560A DC load we're working with above. Here is my "+Temp Rise" or ΔT formula, after T1 + T2 + derating + continuous adjustments, after dividing Tot-I by parallels x 2:

30c + (90c - 30c) * (280A)^2 / (350A * 0.8)^2 = 90c

 

[Smart $]

Please check my 90c operating results for this 560A DC load we're working with above. Here is my "+Temp Rise" or formula, after T1 + T2 + derating + continuous adjustments, after dividing Tot-I by parallels x 2:

30c + (90c - 30c) * (280A)^2 / (350A * 0.8)^2 = 90c

Your calculation looks good for a continuous load, but let me iterate that your result 90?C = T2, and Temperature Rise (ΔT) = 60?C. However, I do not see any adjustment for non-continuous loading in your calculation, for (350A * 0.8) appears to me as derating. I am uncertain how to implement such an adjustment.

Aside from the above, do not forget if this is an underground run, the ambient temperature is lower.

 

[Smart $]

My use of term "burn up" refers to equipment listings...

So a full 560A load operates at 90c, if we count all DC conductors as CCC's. The positive polarity is two cables + the negative polarity is two cables, for a total of four CCC's.

You have to remember your calculated temperature is of the conductor inside conduit. Terminations are not in conduit and heat is better dissipated in a volume less constricted, as in a panelboard. By this reasoning the temperature of a conductor could be 90? in the conduit and lower at the terminals. Whether it is 75? or less would be the question, and I possess no empirical evidence one way or the other.

 

[don_resqcapt19]

My use of term "burn up" refers to equipment listings. I assume most lugs, boxes, & enclosures are not listed for 90c, much less for DC rated for continuous use at 124c.

If we don't agree on 90c as the operating temp., then lets look at my derating first.
Two #350's = 2 x 350A @ 90c or (700A * 0.8) with four CCC's = 560A @ 90c.

So a full 560A load operates at 90c, if we count all DC conductors as CCC's. The positive polarity is two cables + the negative polarity is two cables, for a total of four CCC's.

The derating rules say we can use the derated ampacity on a 75?C rated device as long as the load does not exeed that permitted for a 75?C conductor. If what you say is correct, the derating tables are not correct.

 

[ramsy]

The derating rules say we can use the derated ampacity on a 75?C rated device as long as the load does not exeed that permitted for a 75?C conductor. If what you say is correct, the derating tables are not correct.

Yes, I see it. My 90c calc for T2 exceeds what code calls a permissalbe termination for #350 @75c.

However, practicing this code minimum can conflict with 2005 NEC Examples, which illustrate derating rules that do not permit "worse case".

...124,000 VA / (480V ? 3) = 149 A, or a 1 AWG conductor @ 90?C) could not be used because the termination result (1/0 AWG based on the 75?C column of Table 310.16) would become the worst case, requiring the larger conductor.

In this NEC example the load is 149A, and table 310.16 @75c allows 150A (all good), but this "worst case" requires the next larger conductor @ 175A (why).

IMHO tracking T2 for terminaitons, along with ampacity & derating, clarifies this derating-code minimum, but my T2 model will require some verrified and validated examples, or help from private authorities (ieee), etc..

Any pointers to help get my T2 down from 90c to 75c would be greatly appreciated.

 

[Smart $]

Yes, I see it. My 90c calc for T2 exceeds what code calls a permissalbe termination for #350 @75c...

Not really.

As I said above, your T2 calculation is for conductor temperature in a conduit... and only an approximate one at that.

Terminations do not occur in conduit...​

...but rather generally occur in a space having more volume and better heat dissipation.

An analogous situation would be using a small space heater to heat a single, small room versus the same heater used to heat a large, multi-room house. It is therefore not unrealistic to believe a conductor's temperature could be <75?C at its terminations and a lot closer to 90?C somewhere along its run inside a conduiit.

 

[dnem]

Your calculation looks good for a continuous load, but .....

Houstonthathasaproblem checked out a long time ago without answering the over or under 560 amp load question. . I wouldn't even worry about continous load until he finds out what his load is.

If his load is continuous then 215.2(A)(1) + 230.42(A) both say that you multiple the load by 125% "before the application of any adjustment or correction factors".

But I think Houstonandhisproblem are gone for good because we kept letting the code get in the way of the answer that he wanted.

 

[Smart $]

Houstonthathasaproblem checked out...

Good thing this isn't a critical mission :grin:

If his load is continuous then 215.2(A)(1) + 230.42(A) both say that you multiple the load by 125% "before the application of any adjustment or correction factors".

OK... so let's say he actually has a calculated load of 600A... 500A continuous (@125%) and 100A non-continuous (@100%). Would the existing installation be compliant without any beefing up?

 

[ramsy]

Houstonthathasaproblem checked out a long time ago .. I wouldn't even worry about continous load ...gone for good because we kept letting the code get in the way of the answer that he wanted.

The temp rise issue we've been working on for several years is related to this OP, and Smart's help can't be waisted now that we've got him in one place.

Terminations do not occur in conduit...​

...but rather generally occur in a space having more volume and better heat dissipation. ..It is therefore not unrealistic to believe a conductor's temperature could be <75?C at its terminations and a lot closer to 90?C somewhere along its run inside a conduiit.

Yes, terminations occur in exclosures, however, any hypothesis needs a test. The chart I found for enclosure heat dissipation is a temp. rise curve that only goes in one direction; up. It appears enclosures always cause conductors to get hotter to some degree, but never cooler. Please seen "Heat Dissipation in Electrical Enclosures"
http://www.westernextralite.com/resources.asp?key=72

 

[ramsy]

As I said above, your T2 calculation is for conductor temperature in a conduit...

I thought conductor lengths were refered to as heat sinks emerging at the same temp. from conduit for several feet, unless exposed to flash freezing or refridgeration.

Regardless, of our different perspectives on this enclosure termination temperature, we should agree that some reliable examples of conductor temperature calcs for terminations would help solve this issue.

The 2005 NFPA 70 handbook is one source of refined examples for voltage drop calcs, using Neher-McGrath after the Table 9 section in the handbook, so I'm happy with my VD calcs, but I was hoping someone with a 2008 NFPA 70 handbook might find similar examples for termination temperatures, under the Neher and M. H. McGrath section in either 310.15(C) or Table 9 section of the 2008 handbook.

 

[ramsy]

Our last reference to this IEEE temp. rise formula follows the thread below.
http://forums.mikeholt.com/showpost.php?p=639473&postcount=35

My 90c operating-temp calc below for the OP's 560A DC load does not include continuous adjustments..

T1 + (TR-T1) * Load^2 / Imax^2 = T2
30c + (90c - 30c) * (280A)^2 / (350A * 0.8)^2 = 90c

Factoring the 280A load x 1.25 for continuous use results in 124c.

 

[Smart $]

...
My 90c operating-temp calc below for the OP's 560A DC load does not include continuous adjustments..

Oh, but it does!!! You are using 280A for the load and the ampacity of each conductor. This value includes 125% of the continuous loads.

First, let's use the example I posed for David:

600A (2?300A) calculated load
500A (2?250A) @ 125% of continuous loads
400A (2?200A) @ 100% of continuous loads
100A (2?50A) @ 100% of non-continuous loads​

So when you determine T2 you should only use 100% of continuous loads.

30c + (90c - 30c) * (400A + 100A)^2 / (350A * 0.8)^2 = 78c​

Looking at a 280A calculated load and assuming it is all continuous, we get:

30c + (90c - 30c) * (280A * 0.8)^2 / (350A * 0.8)^2 = 69c​

The odd thing that happens in calculating this way is demonstrated by a full 280A non-continuous load.

30c + (90c - 30c) * (280A)^2 / (350A * 0.8)^2 = 90c​

I have yet to determine any simple way around this anomaly. But as you can see, we are still in the ballpark of required values when the majority of the load is continuous. There are way too many variables to accurately calculate terminal temperature with a simple single formula. This is exactly why NEC requirements on the matter are as they are. Additionally, you have to remember this formula only reasonably approximates temperature rise, and was developed for use in voltage drop calculations, an issue the NEC does not make binding.

 

[Smart $]

Yes, terminations occur in exclosures, however, any hypothesis needs a test. The chart I found for enclosure heat dissipation is a temp. rise curve that only goes in one direction; up. It appears enclosures always cause conductors to get hotter to some degree, but never cooler.

Just the fact that current passes through an enclosure will make it get warmer... I?R loss. Cut back on the current and the temperature will drop.

 

[Smart $]

I thought conductor lengths were refered to as heat sinks emerging at the same temp. from conduit for several feet, unless exposed to flash freezing or refridgeration.

Perhaps... but that does not mean the conductor where it emerges is the same temperature as it is a several feet into the conduit either. Temperature of the conductor concerns all locations along its entire length. Heat sinking works in all available directions, same as current. Because the metal mass centralizes at panelboards and such, I assume the heat sink effect is outward, not inward.

Regardless, of our different perspectives on this enclosure termination temperature, we should agree that some reliable examples of conductor temperature calcs for terminations would help solve this issue.

Without a doubt... or perhaps even some good empirical data.

 

[petersonra]

I need to increase the ampacity of a 250VDC feeder to 600amps. The existing feeder utilizes parallel 350MCM (two conductors per leg) THHN in a 3 inch EMT conduit which is good for around 500amps after derating (using the 75 degree C rating of the conductor). In order to meet the required ampacity of 600amps is it permissible to run another pair of 350MCM conductors alongside of this existing installation within a separate raceway? The feeder is well over 500ft. long and I am trying to avoid pulling 6 new (#4/0) cables.

Is there some reason you can't use the 90 degree column and just use a splice block at the ends that is rated for 90 degree? Would get you close to 600A.

 

[Smart $]

Is there some reason you can't use the 90 degree column and just use a splice block at the ends that is rated for 90 degree? Would get you close to 600A.

That's a good idea!!!

Main conditions would be the run conductors termination and terminal block must be rated 90?C and the "jumpers" would have to be sized larger.

 

[ramsy]

You are using 280A for the load and the ampacity of each conductor. This value includes 125% of the continuous loads. ..So when you determine T2 you should only use 100% of continuous loads.

OK I agree, my derating attempt (350 * 0.8) may use Tbl. 310.15(B)(2)(a) adjustements for now, since other heat sources in same conduit will affect T2.

However, while keeping ampacity ajustments per NEC, my T2 calc is no longer changed by continuous vs non-continuous adjustments, since I2R (heat) losses aren't governed by hours.

Engineering assistance will probably be needed to improve this T2 calc in the future, and may have something to do with replacing my Tbl. 310.15(B)(2)(a) derating adjustements, since IEEE does not use this table nor continuous factors at all for T2.