Large Current Carrying Industrial Wires?

[TransistorGeek]

Greetings!

I have a purely resistive DC load of 325A at 10V in an industrial application. The load is approximately 50 feet from the power supply. I'm not sure how to find the appropriate wire for this size load at this distance to fit this application. The standard AWG table doesn't appear to go to this high current and the 12.4/0, which seems to be the best fit for the amperage based on a maximum 3% voltage drop, is too large and unweildy.

Does the NEC standard [FPN's to 210-19(a), 215-2(b), and 310-15] "...a maximum of 3% voltage drop for branch circuits, a maximum of 3% voltage drop for feeders, but a maximum of 5% voltage drop overall for branch circuits and feeders combined" even apply to industrial equipment?

Here are my calculations...

Single-phase DC Load:

(Conductor Resistivity)(2)(Amps)(Distance in Feet) = Wire Circular Mils
(Allowable Voltage Drop)

Conductor Resistivity = Copper; 11.2
Amps = 325A + (325A*0.2) = 390A
Distance in Feet = 50 ft.
Allowable Voltage Drop = (0.03)*(10V) = 0.3V

Thus,

(11.2)(2)(390)(50) = 1456000 circular mils = 1,456 MCM
0.3

1,456 MCM ~ 12.4/0 wire

This wire is way too huge to bend and fit in our application. Even if I don't size for 80% of the load, and use 325A in my calculation instead of 390A, I still get approx. 1,214 MCM, which is also too large. So my question is whether the 3% maximum voltage drop is really applicable in this industrial application, and how to calculate the acceptable voltage drops.

If, for example, a 5V drop would be acceptable, we could simply size our power supply for the extra 5V and go with a 1/0 wire, which is far more manageable. The heat generated by the power loss doesn't appear to be worrisome:

P_Loss = (I^2)*R

R = p(L/A) where p = resistivity of copper, L= length (m), A = area (m^2)

so,

Resistivity of copper = (1.69*10^-8)
Length = 50 ft. = 15.24 meters
Area = 0.0534705 m^2

(1.69*10^-8)[(15.24)/(0.0534705)] = ~ 4.82 uOhms (or 4.82*10^(-6))

Therefore,

(325A)^2*(4.82uOhms) = ~0.509W (power lost as heat)


So would this scenario, having a 50% voltage drop, be acceptable or would it be in violation of the NEC, UL, or IEC standard(s)? If it would be a violation, why? How can we calculate the temperatures expected as a function of time, given ambient air conditions, based on this power loss as heat?

Some might suggest rating the power supply for a much larger voltage and simply using a step-down transformer at the load, but we don't have sufficient space at the load for this to be feasible.

My primary goal here is to find a proper solution, but I'd also like to understand more about the reasoning behind it as well, so please don't be shy about getting into detail. The more information you can share the better! Thank you all in advance for your assistance!

 

[charlie b]

I don't understand your notation:

1,456 MCM ~ 12.4/0 wire

There is #12 AWG wire. There is #4/0 wire. But what is 12.4/0? If you mean to say 12 wires in parallel, each being a #4/0, then the math does not work out for me.

But the real answer to your question is that (1) The 3% and 5% numbers are essentially suggestions, not requirements, and that (2) The question of what is, and what is not acceptable boils down to what the load can tolerate as its voltage supply level.

 

[GoldDigger]

With DC, a stepdown transformer is not an option.
Putting the DC supply itself closer would be.

Tapatalk...

 

[LEO2854]

Welcome aboard:thumbsup:

As noted the informational notes are not mandatory look at 90.5 in the 2011 NEC.

90.5 Mandatory Rules, Permissive Rules, and Explanatory Material.(A) Mandatory Rules. Mandatory rules of this Code are those that identify actions that are specifically required or prohibited and are characterized by the use of the terms shall or shall not.
(B) Permissive Rules. Permissive rules of this Code are those that identify actions that are allowed but not required, are normally used to describe options or alternative methods, and are characterized by the use of the terms shall be permitted or shall not be required.
(C) Explanatory Material. Explanatory material, such as references to other standards, references to related sections of this Code, or information related to a Code rule, is included in this Code in the form of informational notes. Such notes are informational only and are not enforceable as requirements of this Code.
Brackets containing section references to another NFPA document are for informational purposes only and are provided as a guide to indicate the source of the extracted text. These bracketed references immediately follow the extracted text.
Informational Note The format and language used in this Code follows guidelines established by NFPA and published in the NEC Style Manual. Copies of this manual can be obtained from NFPA.
(D) Informative Annexes. Nonmandatory information relative to the use of the NEC is provided in informative annexes. Informative annexes are not part of the enforceable requirements of the NEC, but are included for information purposes only.

 

[ggunn]

Greetings!

I have a purely resistive DC load of 325A at 10V in an industrial application. The load is approximately 50 feet from the power supply. I'm not sure how to find the appropriate wire for this size load at this distance to fit this application. The standard AWG table doesn't appear to go to this high current and the 12.4/0, which seems to be the best fit for the amperage based on a maximum 3% voltage drop, is too large and unweildy.

From the Southwire voltage drop calculator (http://www.southwire.com/support/voltage-drop-calculator.htm):

"4 conductors per phase utilizing a #750 Copper conductor will limit the voltage drop to 2.74% or less when supplying 325.0 amps for 50 feet on a 10 volt system."

That's a lot of copper.

 

[petersonra]

how did you determine the wire size was "too large" if the calculation says it has to be that big for the VD you want and the length of the conductors?

personally, I would be moving the power supply a lot closer as suggested by another poster.

if you went with 1/0 I think the wire would get really warm. will probably exceed the temperature limits allowed for the insulation and damage the insulation for normal wire. maybe if you used teflon coated wire it might work.

1/0 is 0.096 Ohms/1000 feet

I^2*R = > 1000 W

many power supplies have remote sense built into them to accommodate this kind of thing. if you have remote sense and the power supply can accommodate the voltage drop, I would be inclined to have at least (3) 1/0 wires.

 

[GoldDigger]

Keep in mind that unless the load current is variable and the load is voltage sensitive, you may just want to accept the higher percentage voltage drop and raise the supply voltage accordingly.
Even if you cannot move the power supply closer, you may be able to configure remote voltage sensing instead. That will not reduce the wasted power but it will assure proper equipment performance.

What is the load?


Tapatalk...

 

[TransistorGeek]

charlie b,

Thanks for the reply. I don't mean to say multiple wires in parallel. The 12.4/0 is from the "Aught" entries which, unless I'm mistaken, are a less commonly used continuation of the AWG table. Wikipedia refers to these as "Fictitious AWG equivalents." Here are a few quick resources one might find from googling "12.4/0"

Source 1

Source 2

Source 3 (Wikipedia)


The load needs 10V, period. Getting the correct voltage to the load isn't so much the problem as ensuring it can be delivered safely and without violating any codes. If a 33% (corrected) voltage drop is acceptable, and the heat generated by 509 mW doesn't exceed the rating of the insulation, then we should be golden. It sounds like a voltage drop limitation isn't defined by any standards, but only recommended. If this is the case then the only concern is the heat that might be generated by power loss. Does anyone recall how to perform this calculation? My thermodynamics is rusty to say the least... Once I know the maximum temperature increase due to power lost as heat, I'll be able to search for the appropriately rated insulation. If no insulations exist for the temperatures calculated, then I'll need to go larger than 1/0 wire size until I find one with insulation capable of handling the corresponding temperatures. The temperatures will decrease as the wire sizes increase. Do you agree?

 

[LEO2854]

This is the table from Source 3 (Wikipedia) ,so if I'm right 12.4/0 = 1500 KCMIL conductor.

& solid copper equivalents
kcmil
MCM mm? Diameter Fictitious AWG
equivalent
in. mm
250 126.7 0.500 12.70 4.7/0
300 152.0 0.548 13.91 5.5/0
350 177.3 0.592 15.03 6.2/0
400 202.7 0.632 16.06 6.7/0
500 253.4 0.707 17.96 7.7/0
600 304.0 0.775 19.67 8.5/0
700 354.7 0.837 21.25 9.2/0
750 380.0 0.866 22.00 9.5/0
800 405.4 0.894 22.72 9.7/0
900 456.0 0.949 24.10 10.2/0
1000 506.7 1.000 25.40 10.7/0
1250 633.4 1.118 28.40 11.7/0
1500 760.1 1.225 31.11 12.4/0
1750 886.7 1.323 33.60 13.1/0
2000 1013.4 1.414 35.92 13.7/0

 

[Besoeker]

My primary goal here is to find a proper solution, but I'd also like to understand more about the reasoning behind it as well, so please don't be shy about getting into detail. The more information you can share the better! Thank you all in advance for your assistance!

Greetings to you to!!

As most here probably know, I'm from across the pond and we do metric (SI) rather than Imperial.
At a first stab I'd go with two 95mm^2 for the pos and for the neg if you want something not too difficult to handle.
That's about 3/0 in your money.
Voltage drop for the round trip would be about 2.5V.
I think you will be stuck with something like that if you don't want to hugely oversize the conductors than they need to be for current carrying.

 

[TransistorGeek]

Wow, thanks for the replies! I believe I found my new favorite forum!

We can't move the power supply closer to the load. The load is an electrolytic cell. The voltage needs to be fairly accurate at 10V +/- approximately 5%. I'll research the remote sense feature as this is something I'm currently not familiar with utilizing. I do like the idea of using parallel cables at smaller gauges though, but this may present a new problem based on space requirements.

 

[Besoeker]

Wow, thanks for the replies! I believe I found my new favorite forum!

We can't move the power supply closer to the load. The load is an electrolytic cell. The voltage needs to be fairly accurate at 10V +/- approximately 5%. I'll research the remote sense feature as this is something I'm currently not familiar with utilizing. I do like the idea of using parallel cables at smaller gauges though, but this may present a new problem based on space requirements.

Is it a controlled rectifier producing the 10Vdc?
Sensing the voltage at the load end and feeding it back to the controller shouldn't be a problem over such a relatively short distance.
We'd use two core screened.

 

[TransistorGeek]

It is a custom power supply for this contract. I have unfortunately not been involved in the power supply specification process, but I'm sure it will be a regulated, switching power supply. Whether it will have remote sensing capability or not is currently a question I am researching... It looks like the mechanical engineers put together the power supply purchase spec, so I'll be having some fun going through it for some time. :happyno:

 

[zbang]

Thanks for the reply. I don't mean to say multiple wires in parallel. The 12.4/0 is from the "Aught" entries which, unless I'm mistaken, are a less commonly used continuation of the AWG table. Wikipedia refers to these as "Fictitious AWG equivalents."

Since you're less familiar with the trade usage, once we hit 4/0 size, we start using the MCM numbers instead. So 3/0, 4/0, 250mcm, 300mcm, etc. That's what confused some folks here.

It's still a big expensive, hard to handle, wire.

 

[Jraef]

Since you're less familiar with the trade usage, once we hit 4/0 size, we start using the MCM numbers instead. So 3/0, 4/0, 250mcm, 300mcm, etc. That's what confused some folks here.

It's still a big expensive, hard to handle, wire.

And just to be a little pedantic, we no longer use "MCM" any more either, at least not officially. It is now "kcmil", as in "1000 circular mils". The first "M" in MCM was always wrong according to some. Roman numeral M is 1000, which is where it came from, but as an SI measurement unit, 1000 is abbreviated as "k", because "M" means "1/1000". "Mils" was already a reference to 1/1000 of an inch, so the term "MCM" was immediately contradictory if you think about it, which of course lots of people did.

 

[zbang]

And just to be a little pedantic, we no longer use "MCM" any more either, at least not officially.

Old habits and all that . And to further muddy the waters, MCM (Roman) = 1900 (decimal), neither of which relates to the actual wire size.

 

[Besoeker]

And just to be a little pedantic, we no longer use "MCM" any more either, at least not officially. It is now "kcmil", as in "1000 circular mils". The first "M" in MCM was always wrong according to some. Roman numeral M is 1000, which is where it came from, but as an SI measurement unit, 1000 is abbreviated as "k", because "M" means "1/1000".

To be a little more pedantic, M means million as in MW. Megawatt.
Lower case m means 1/1000 as in mW. Milliwatt.
Usually, the context makes it clear which is intended even if the wrong prefix is used.