Service Cable Calculation

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Michael15956

Senior Member
Location
NE Ohio
How does one calculate the circular mills of Service Cable?

Chapter 9, Note (9) states that the cross-sectional area must be used by using the major diameter of the ellipse as a circle diameter.

Should I just measure the ellipse and use Table 8, Conductor Properties, column for diameter?


TIA
 

chris kennedy

Senior Member
Location
Miami Fla.
Occupation
60 yr old tool twisting electrician
Michael15956 said:
How does one calculate the circular mills of Service Cable?

Chapter 9, Note (9) states that the cross-sectional area must be used by using the major diameter of the ellipse as a circle diameter.

Should I just measure the ellipse and use Table 8, Conductor Properties, column for diameter?


TIA

I don't get the question. If its 2/0 SE then look at 2/0 =133100 cir.mils.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
If I understand the question, Michael is asking how to calculate the cross sectional area of a _cable_, including several conductors, their insulation, and the over-all sheath, for purpose of calculating conduit fill.

For conduit fill purposes, the requirement for cables is that you treat them as a single large conductor. So if you had a cable with 5 conductors and a diameter of 3/4 inch, for purpose of conduit fill you would treat it as a single conductor 0.75 inch in diameter. As a _single_ conductor, it could fill 53% of the conduit, rather than using the normal 40% fill limit for 'more than 2' conductors.

There is an additional requirement for 'cables that have an elliptical cross section'. This requirement is that you find the major (largest) diameter of the ellipse, calculate the area of the _circle_ with that diameter, and then use the area of the _circle_ as the cross section of the cable for purposes of conduit fill.

The formula for the area of a circle is simply A = pi*r^2 . However if you are working in 'circular mils', then the formula is even simpler. The unit 'circular mil' already takes into account the pi factor.

Given a diameter in mils (0.001 inch), the area of a circle in circular mils is simply the diameter squared. Thus if you have a cable with a major diameter of 0.307 inches, you must treat is as having a circular cross section of 307^2 = 94249 circular mils.

As an aside, the requirement in note 9 applies to cables with an _elliptical_ cross section. An ellipse is a very specific shape, and most cables are _not_ ellipses...IMHO the requirement should probably be applied to any non-circular cable :)

-Jon
 

nakulak

Senior Member
or sometimes its easiest to try to find the cable on the manufacturer's website and look up the data sheet for the info
 

Michael15956

Senior Member
Location
NE Ohio
winnie said:
If I understand the question, Michael is asking how to calculate the cross sectional area of a _cable_, including several conductors, their insulation, and the over-all sheath, for purpose of calculating conduit fill.


-Jon

Jon,
You are correct, the above was the intent of my post. But, I cannot follow your calculations.
 

mdshunk

Senior Member
Location
Right here.
I take a little different approach to this, since you never know which brand of cable the supply house will supply on any given day. I measured the gland on the weatherproof SE connectors, since they are sized to accomodate all of today's modern SE cable. For 200 amp SE, I come up with a major diameter of 1.440". For 100 amp SE cable, I come up with a major diameter of exactly 1".

sediameter1.jpg


sediameter2.jpg


sediameter3.jpg
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Michael15956 said:
Jon,
You are correct, the above was the intent of my post. But, I cannot follow your calculations.

The area of a circle is the constant 'Pi' times the square of the radius. The radius is half of the diameter. This equation works for _any_ linear unit. If you measure your radius in inches, then the answer is given in square inches. If you measure your radius in miles, then the answer is in square miles.

So for a circle 3" in diameter, the area is Pi * 1.5 ^ 2, which equals Pi * 1.5 * 1.5, which equals 3.14159265... * 1.5 * 1.5 = 7.06858...square inches.

The unit 'circular mil' has the Pi constant built in, making the equation simpler. The area in circular mil is simply the square of the diameter in mil. If you have a circle 3 mils in diameter, then its area is simply 3^2 = 9 circular mils.

The other thing that you need to know is that a mil is 1/1000 inch. So if you have a circle 3" in diameter, it has a diameter of 3000 mil, and an area of 3000 ^ 2 = 9000000 circular mil.

Taking mdshunk's numbers, you take the '100A SE' which has a major diameter of 1". This equates to 1000 mil, thus you need to treat is as a single conductor with an area of 1000 ^ 2 = 1000000 circular mil = 1000 kcmil.

-Jon
 

Michael15956

Senior Member
Location
NE Ohio
Thanks much Marc I like how you did that!

Jon,

Is this right for Marc's 200amp example?

1.440 * 1000 equates to 1,444 mils
1,444 ^ 2 equates to cmils 2,073,600
2,073,600 cmils equates to 2,073.6 kcmils
 

Michael15956

Senior Member
Location
NE Ohio
Marc,
So, for your example of 100amp SE Cable at 1" would you install 2" conduit at 53% fill?

200amp SE Cable at 1.440 install 3" conduit at 53% fill?
 

mdshunk

Senior Member
Location
Right here.
Michael15956 said:
Marc,
So, for your example of 100amp SE Cable at 1" would you install 2" conduit at 53% fill?

200amp SE Cable at 1.440 install 3" conduit at 53% fill?
Without doing the math myself, I guess that sounds about right. I really wouldn't put SE in a raceway anyhow, but I know some people do. I've seen it done where they run SE inside a meter's line side mast, then use unprotected SE for the meter's load side. For 3" pipe, you're not really going to find a 200 amp meter can with a 3" hub, since 2-1/2" is the biggest size most brands of 200 amp meter cans will accomodate. On the load side, you can't even cut a 3" knockout, since there's not enough depth to do so considering that the lid overlaps the bottom a bit. The underground fed type of meter can will accomodate a 3" pipe, but it's really up in the air (for me at least) whether SE is suitable for pulling in pipe underground. It's a dumb application for that type of cable, in the least.
 

Michael15956

Senior Member
Location
NE Ohio
Thanks Marc,

I really wouldn't put SE in a raceway anyhow, but I know some people do. I've seen it done where they run SE inside a meter's line side mast, then use unprotected SE for the meter's load side.

This is also what I saw another electrician do. When asked why, he said it was easier than pulling three wires. Although he did install 2" PVC for the installation of 100amps.
 

mdshunk

Senior Member
Location
Right here.
Michael15956 said:
This is also what I saw another electrician do. When asked why, he said it was easier than pulling three wires. Although he did install 2" PVC for the installation of 100amps.
It probably is a bit easier, and also a good way to use up the shorter scraps of SE cable.
 

Bob NH

Senior Member
Michael15956 said:
Marc,
So, for your example of 100amp SE Cable at 1" would you install 2" conduit at 53% fill?

200amp SE Cable at 1.440 install 3" conduit at 53% fill?

Not correct.

For a single cable in a conduit, the area of the cable must not exceed 53% of the area of the conduit. Since the area of the conduit and the area of the cable are proportional to the diameter squared, the diameter of a single cable must not exceed the square root of the inside diameter of the conduit.

The square root of 0.53 is 0.728. Therefore, a single cable may be as large as 0.728 times the diameter of the conduit.

The other way of calculating it is that the diameter of the conduit must be at least 1/0.728=1.3736. Therefore, a 1" diameter cable could be legally installed in a conduit that is at least 1.3736" times the diameter of the cable. A 1 1/2" conduit would meet that requirement.

A 1.440 diameter cable could be installed in a conduit that is 1.440/0.728 = 1.978" which is a 2" conduit.
 
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Michael15956

Senior Member
Location
NE Ohio
Bob NH said:
Not correct.

For a single cable in a conduit, the area of the cable must not exceed 53% of the area of the conduit. Since the area of the conduit and the area of the cable are proportional to the diameter squared, the diameter of a single cable must not exceed the square root of the inside diameter of the conduit.

The square root of 0.53 is 0.728. Therefore, a single cable may be as large as 0.728 times the diameter of the conduit.

The other way of calculating it is that the diameter of the conduit must be at least 1/0.728=1.3736. Therefore, a 1" diameter cable could be legally installed in a conduit that is at least 1.3736" times the diameter of the cable. A 1 1/2" conduit would meet that requirement.

A 1.440 diameter cable could be installed in a conduit that is 1.440/0.728 = 1.978" which is a 2" conduit.

Bob,

Thanks a lot. This I understand and I can use it in the future!
 

tryinghard

Senior Member
Location
California
Maybe this has already been said, and if so feel free to tell me to shut up, but here?s what I use for cable:
in square = Diameter square x (pie/4), once you have in square simply go to chapter 9 table 4 and use 53% column for single cables and 40% column for more than 2 cables.
 
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