Conductors for General Use

[jsinclair]

This is a simple question that I should know the answer to. However this DC work is new to me. Are the tables in 310 applicable to both AC and DC? Is the NEC in general applicable to both AC and DC? I am pretty sure that it is, with its exceptions of course, but I can't find anything within the code that solidify's this. Thanks for any help in advance.

John

 

[lrollo]

Re: Conductors for General Use

In my opinion, as long as the voltage doesnt exceed the insulation rating, such as 300 volts or 600 volts, and you did your calculations properly to determine the amps and sized the condutor to that size, I think the wire chart on 310.16 will work for AC or DC.

 

[jsinclair]

Re: Conductors for General Use

Got time for some more information now. (This is actually really bothering me that I can't come to some conclusion.) Here's the data:

24 vdc power supply
50' wire length (one way)
80 load
.5v target voltage drop
12.9 K value

CM = (2 x K x L x I)/VD

CM = (2 x 12.9 x 100 x 80)/.5 = 412800 cmils

If I am doing this correct, then I interpret this setup as requiring a 500mcm cable to maintain a .5v drop an 80A load over a 50' distance. Does anyone concur?

 

[earlydean]

Re: Conductors for General Use

The formula cm = 2 x I x k x L / VD is already figured for the return path. L = the distance one way to the load. Your cm answer is twice as big, and the wire should be 250 kcmil. But why do you need to limit the voltage drop to 1/2 volt? Three percent of 24 is .72 or three quarters of a volt. Makes the wire 3/0. That won't really help. Why not start with a somewhat higher voltage, if it is that important? If it is a motor it is not, as the speed will be less, so crank it up.

 

[dereckbc]

Re: Conductors for General Use

jsinclair, I design 48 and 24 VDC battery plants and the distribution systems. Voltage drop is the most important factor in these systems. You can use 310.16 for minimum cable sizes, but I do not use it other than a safety check. I use voltage drop. In battery plants, the batteries in most cases do not have OCPD's, so 310.16 is of no use.

I use the following formula:
CM=(22.2 * I * D)/VD
Where
CM = Circular Mills
I = Max load current
D = One way cable distance
VD = Voltage Drop

I use 22.2 which is equal to 2 * 11.1 (K for copper) I find 12.9 is too high a value, 11.1 is used as an industry standard.

Voltage drop is the tricky part. 2 volts is the standard for 48 VDC, and 1 for 24 VDC. What makes it tricky is the distribution system is divided into separate sections.

For example; a typical 48 VDC battery plant you have three sections of distribution.
1. Batteries to Charge Buss = .25 VDC
2. Main Power Board (primary feeder distribution) to Breaker Bay (secondary branch distribution) = .5 VDC
3. Breaker Bay to equipment loads (branch circuits) = 1.25 VDC

For a 24 VDC plant these values are cut in half. If you use the voltage drop method the conductors will always be larger than 310.16 requirements unless you are using very short distances. So always double check.

For example in breaker bays the smallest breaker we use is a 20-amp breaker, and the smallest conductor we use with it is a 6 AWG.

That's about it except for grounding, but that is a subject for another day.

 

[dereckbc]

Re: Conductors for General Use

Originally posted by jsinclair:
24 vdc power supply
50' wire length (one way)
80 load
.5v target voltage drop
12.9 K value

CM = (2 x K x L x I)/VD

CM = (2 x 12.9 x 100 x 80)/.5 = 412800 cmils

If I am doing this correct, then I interpret this setup as requiring a 500mcm cable to maintain a .5v drop an 80A load over a 50' distance. Does anyone concur?

Use 22.2 for K, and 50 feet oneway distance. Should get 177600 cm 4/0.

[ January 23, 2004, 05:27 PM: Message edited by: dereckbc ]