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Medium Voltage Underground Cable / Smallest Available Size 3

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saladhawks

Electrical
Jun 4, 2004
86
I have an application for a 15 kV, 133% EPR Insulation (220 Mil) underground cable that needs to fit in 2" Conduit (1.913" I.D.). I only need an ampacity of 70 Amperes, but have been told by my local Okonite rep. that they will not manf. a medium voltage cable smaller than #2 CU per NEC Table 310.5 (2002 Version). Note that this is an utility application and we are not under obligation to meet NEC requirements. I checked the 2007 NESC and was not able to locate a minimum UG medium voltage cable limit. Does anyone know of a UG cable manufacturer who manufacturers cables smaller than #2 CU?
 
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The minimum conductor size for a MV cable is limited by the peak voltage stress, which occurs near the conductor shield. #2 is the smallest that I've seen in manufacturers' catalogs. It doesn't matter if it's aluminum or copper conductor.

You might want to ask your Okonite rep what the peak stress would be for #2 (which they will make) or #4 (which they won't). I'm sure you'll see a dramatic difference in peak stress and I believe that is why they won't make a smaller size than #2 at 15 kV.
 
saladhawks,

Like magoo2 said...

It's not an NEC issue. The cables are manufactured in accordance with AEIC and ICEA standards; max electrical stress determines the minimum conductor diameter for a voltage class (15kv - #2, 25kv - #1, 35kv - #1/0, eg).
 
Yes, Keith. It's a nonlinear electric field. The formula for calculating the peak stress in volts/mil is:

S = 0.868 E/[d_c log10 (D/d_c)]

E = l-g voltage
D = diameter of insulation in mils (under insulation shield)
d_c = diameter of conductor in mils

d_c - supposed to be d with the subscript c

See Southwire Power Cable Manual.
 

For cables I have always seen that the maximum electric field occured at:

Emax= V/R2ln(R1/R2)

Where R2 is the conductor radius from center and R1 is the radius to the outside of the insulation. This equation holds true for just a conductor surrounded by insulation with no Intersheath or shield.

A cable havin an intersheath or shield has its maximum field represented by:

Emax= (V/R3ln_a)(1/1+a)

a = R1/R2 = R2/R3

Where R3 is the radius of the conductor and R2 is the radius to the shield or sheathe and R1 is the radius to the outside of the insulation.

I belive that the shield or sheathe is used to reduce the maximum electric field in cables and can be seen by examles with the above equations.

At least this is how I understood it. Please correct me if I am wrong.
 
I believe your first equation and the one I got from the Southwire manual are equivalent.

log10 x = (ln x)/2.303 = 0.4343 ln x

mine used the diameter and yours used the radius.

With these 2 substitutions, they match.

I didn't check your other eqn.
 

I have seen these equations and understand them in theory but never really had an explanation for how they are used in cable applications.

I am curious now, is the reason that an MV cable has a shield for the purpose of reducing the maximum electric field in a conductor. From the second equation I posted you can do examples which show the electric field diminishing by having this inner sheathe. Is this the reason MV cable is shielded?
 
I think a picture would help and I'll try to scan one tomorrow.

The shield is intended to make the dielectric field (i.e., the electric field that occurs across the insulation) more controlled and more symmetrical.

I think the wear on an automobile tire is a good analogy. If I were to make a tire in a square rather than a round configuration, you'd get nonuniform wear (and a pretty bumpy ride), and the tire wouldn't last as long.

Without a shield, the stress on the inside of the cable is pretty nonuniform and the wear would be much higher. Conductors that are insulated for operating voltages above 2 kV are shielded (5 kV cable and higher). Below that, there's more insulation added to the nonshielded design.
 
That's a good one, jimbofitz.

Here's another one, courtesy of EPRI. It shows the electric field in Fig. 4-3 and the stress distribution on Fig. 4-4.

 
If the smallest MV Underground Cable that can meet voltage stress requirements is #2 (CU or AL), why does the NEC 2002 go through the trouble of listing ampacities of cables all the way down to #8 in Tables 310.67 to 310.86. Looks like a big waste of time to me.
 
I think you missed the point that we were talking about 15 kV cable for the #2 size. You can use #8 for unshielded designs up to 5 kV.
 
NEC 2002 Tables 310.67 to 310.86 show #6 cables for voltages up to 35 KV.
 
Several cable Vendors have cables in their marine product lines that seem to meet your requirements and have smaller gauge wires
 
Ask Okonite about their line of X-ray cable. I am not sure that it is up to the NEC but the cable is 220mil EPR with tiny conductors. However, the conductors may be too small for your application. You will also have trouble finding a termination for this cable. If you need more help let me know.

Benjamin Lanz
Past Chair of IEEE 400
Sr. Application Engineer
IMCORP- Power Cable Reliability Consultants
 
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