How do you terminate 750 Kcmil fine stranded cable with 1080 strands using NEC listed lugs made for 61 strands?
Gerald Newton
spec'ed flexible power cable that is UL listed with RHH insulation but
there are 1080 strands and the diameter is about 1/10th of an inch
larger than that listed for 61 strands in Table 8 Chapter 9 of the
NEC. The problem is regular 750 Kcmil lugs are too small. This
leaves two options use a 800 Kcmil lug which the contractor did or cut
a row of strands from the outside of the conductor where it goes into
the lug. So what is the solution? Also does anyone know the formula
for finding the diameter of a cable as a function of the number of
strands? This is an interesting math problem, but I have never seen
the solution.
Don Young
"Gerald Newton" <Gerald...@hotmail.com> wrote in message
news:fc6b0bcb-f468-4132...@q28g2000prh.googlegroups.com...
don't think the nominal size of the lug has any bearing. Lots of cables have
been trimmed to fit smaller connections and I image 99.9% of them are still
working satisfactorily.
I would not trust any formula to accurately determine cable diameter since
there are variables that are not known, such as the tightness and angle of
the twist.
Don Young
Gerald Newton
> "Gerald Newton" <GeraldCNew...@hotmail.com> wrote in message
The formula I am looking for is to find the diameter of a circle that
inscribes a known number of smaller circles each with the same
diameter.
It appears that with an increase in the number of smaller circles the
greater the ratio is of the circumscribed circle diameter squared to
the sum of the smaller circles diameters squared.
In other words, for a given circular mil area (cma), the diameter of a
circumscribed circle in thousandths of an inch squared (cmil area)
grows as the number of smaller circles increases. The sum of the
smaller circles circular mil areas being equal to the given cma.
Bill Kaszeta / Photovoltaic Resources
The correct lugs for these cables are available from Burndy and other
lug manufacturers. They are used all the time in telecommunications.
In the Burndy catalog 'Code' wire is the low strand such as you describe.
There are separate sections for the flex. You will also have to know the
details of the lug attachment hole(s) as to bolt size and spacing if there
are two holes (usual for 750 kcm).
Major flexible cable manufacturers such as Cobra Cable stock these lugs.
Graybar is a major electrical distributor covering most of the USA that
carries some of these lugs.
Once you find a lug part number, try eBay.
Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
bi...@pvri-removethis.biz
Gerald Newton
Resources) wrote:
That is what the engineer is proposing - changing the lugs. But this
demonstrates that when ever products outside the normal NEC are used,
problems arise.
But on the formula. It appears for one strand the circumscribed
circle diameter ratio to the strand diameter would be 1:1. For two
strands the ratio would be 2:1.
For three to infinity strands it appears the ratio would go from 2:1
to 1;1.
The problem is what would the ratio be for three strands, four
strands, five strands.....etc? There must be a mthematical
relationship so the ratios can be calculated.
Bill Shymanski
> On Jul 21, 5:29 pm, "Don Young" <no...@nonesuch.com> wrote:
>> "Gerald Newton" <GeraldCNew...@hotmail.com> wrote in message
>>
>> I would not trust any formula to accurately determine cable diameter since
>> there are variables that are not known, such as the tightness and angle of
>> the twist.
>>
>> Don Young
>
> The formula I am looking for is to find the diameter of a circle that
> inscribes a known number of smaller circles each with the same
> diameter.
> It appears that with an increase in the number of smaller circles the
> greater the ratio is of the circumscribed circle diameter squared to
> the sum of the smaller circles diameters squared.
> In other words, for a given circular mil area (cma), the diameter of a
> circumscribed circle in thousandths of an inch squared (cmil area)
> grows as the number of smaller circles increases. The sum of the
> smaller circles circular mil areas being equal to the given cma.
According to equation 4-13 on page 4-20 of "Standard Handbook for
Electrical Engineers, 11th Edition" edited by Fink and Beatty , the
diameter of the circumscribing circle D, given n layers over the core
wire, each wire of diameter d, is
D = d(2n+k), where k = 1 for a single wire core and 2.155 for a
3-wire core.
I haven't derived this myself (should be possible with a little
high-school algebra) and I hope I haven't made a typo.
Let's check another table in the Handbook: It says a 19-strand #1 AWG
conductor is made up of 19 wires each 0.0664 inches diameter, in 2
layers over the 1-strand core), so k=1, n=2, d=0.0664, and the formula
says D=5d, which is 0.332 inches; and the table says 0.332 inches, so
it's all consistent. The tricky part is that n isn't the number of
wires, but the number of layers of wire over the core - what's the name
for the sequence 1,3,7,19...?
There is also a table on the same page which shows that a stranded cable
is between 15% and 20% bigger in diameter than a solid wire with the
same total copper cross section. The worst case is 12 strands,
everything 19 strands and up is around 15 %.
I can highly recommend this book, I use it regularly, though often just
for the copper wire tables.
Bill
Gerald Newton
>
> - Show quoted text -
Thank you, I will check this out. I used to have an old copy of the
EE handbook; maybe it is still in a box.
daestrom
> On Jul 21, 5:29 pm, "Don Young" <no...@nonesuch.com> wrote:
>> "Gerald Newton" <GeraldCNew...@hotmail.com> wrote in message
>>
>
> The formula I am looking for is to find the diameter of a circle that
> inscribes a known number of smaller circles each with the same
> diameter.
> It appears that with an increase in the number of smaller circles the
> greater the ratio is of the circumscribed circle diameter squared to
> the sum of the smaller circles diameters squared.
> In other words, for a given circular mil area (cma), the diameter of a
> circumscribed circle in thousandths of an inch squared (cmil area)
> grows as the number of smaller circles increases. The sum of the
> smaller circles circular mil areas being equal to the given cma.
I think the densest you can pack circles is in a hexagon packing. So each
strand actually occupies an area of the hexagon that the circular strand
fits in.
If the diameter of a strand is X, then the area of a hexagon that encircles
such a strand is 1.5*X. So if you have 1080 strands of diameter X, the
total area of perfectly packed strands would be about 1620*X. A circle that
has that much area would be 45.4*X. (D= sqrt( 1620*X / (pi/4))
BUT, this doesn't account for skewing of the strands due to the twist. I
suspect that wouldn't add very much though unless you have a short twist.
daestrom
phil-new...@ipal.net
A straight cross section would make for a slight oval shape per strand that is
in twist around the core. And the angle would be greater for the same twist
pitch for more outward conductors. Think of a conductor as wide as the twist
pitch to realize this. Of course the twist pitch would be much greater than
the conductor diameter. That is an interesting geometry problem to consider.
I wrote a program some time back that figured out numbers of hexagonal lattice
points inside a circle for all steps in circle width. I'll have to go find it
and see if I can update it for this kind of geometry. Note that this program
was to figure out lattices kept inside a circle even though the arrangement
was a hexagonal structure. One thing I remember I figured out from it is that
the largest hexagon that would fit inside a circle and would break out of that
circle with any points added was one of a side of 7 points. Larger than that
and you can start adding points at the middle of the sides and still be inside
the circle. But this was done with points, not finite width circles or with
slight ovals (it's purpose was for hexagonal rather than quadrature lattice
data encoding confined well to a circle as opposed to the usual encodings that
stay square).
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