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#### TransistorGeek

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Discussion Starter · ·
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...

(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! :thumbup:

#### McClary’s Electrical

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The NEC fpn on voltage drop is not an enforceable code. It is only a recommendation ( like all fine print notes)

#### KennyW

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Could you not use parallel runs?

#### Bbsound

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I gotta ask, wth uses 300 amps at 10 volts?

#### JRaef

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I gotta ask, wth uses 300 amps at 10 volts?
Lots of stuff. Often times if it is current you need, for example in high temperature electric melt furnaces, or Epitaxial Reactors (a high temp chamber for making silicon wafers) you use a higher voltage and lower current for the longer distance branch circuit, then the voltage is dropped using a transformer so that at the unit to get high current but low voltage.

TransistorGeek,
This is typically done with multiple smaller more manageable cables run in parallel. The rules on applying them are a little more complex because you need to de-rate the cables based on the number of conductors in your raceway etc. etc. Voltage drop is up to you and what your equipment can handle.

#### RIVETER

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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...

(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! :thumbup:
There are so many people on this forum who just try to impress us with their knowledge it is refreshing to hear from someone like you.

#### SteveBayshore

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##### Tool Fetish
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DLO cables installed properly in cable trays. Higher current carrying capacity, a lot more flexible, no derating. Can your equipment operate at a 50% voltage drop?

#### TransistorGeek

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Discussion Starter · ·
Thanks for the kind words, Riveter. The power supply is a custom MIL-SPEC supply designed and built specifically for this application, so technically it could be whatever we want it to be (since we aren't yet locked in a contract). It will be a dedicated supply specifically for an electrolytic cell. The wire(s) or cable delivering power to the load will most likely be run through narrow openings but the needed bend radii are unknown. At the moment I'm merely gathering options, so once more information is known about the specific location, we'll be able to quickly select the best solution.

Thanks again gentlemen! :thumbup:

#### HARRY304E

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There are so many people on this forum who just try to impress us with their knowledge it is refreshing to hear from someone like you.
Well said..:thumbsup:

#### oliquir

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ive seen a lot of similar equipment with welding cable since it is only 10v

#### IMM_Doctor

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No

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?

No, NFPA70 is not a standard. NFPA70 NEC is a code. The scope of NFPA70 is for residential, commercial, and industrial PREMISES (buildings), and are enforceable CODES, by the authority having jurisdiction (AHJ).

You are making a specialized industrial application, and you must engineer the minimums to satisfy the application.

The AHJ will only be able to look at the point of attachment of the power supply from the premises wiring, to the main disconnecting means of your specialized industrial equipment. Your equipment must be labeled to satisfy NEC requirements.

#### ponyboy

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HARRY304E said:
Well said..:thumbsup:
i thought he was being sarcastic

#### Ultrafault

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i thought he was being sarcastic
I think it would have been uncalled for to attack someone for posting there math and reasoning in a profesional forum. So I took it at face value.

#### Jlarson

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DLO cables installed properly in cable trays. Higher current carrying capacity, a lot more flexible

Lots of stuff. Often times if it is current you need, for example in high temperature electric melt furnaces, or Epitaxial Reactors (a high temp chamber for making silicon wafers).
Agree on the DLO, parallel runs of it was often how we did it in semiconductor equipment.

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