Conduit Size (% full)

Conduit Size (% full)

galashkina
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Message 1 of 12

Conduit Size (% full)

galashkina
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I chose Conduit Size - 1’’

The Conduit contains a single wire YEL_14_THHN

 

1.png

 

default.wdw:

 

2.png

 

default.ww1

 

3.png

 

Why is the value of "% full" 79%?

Must be 100%

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Message 2 of 12

Icemanau
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OK, this is from the Help file...

 

<HELP FILE START>

 

The .WDW file contains the wire information. You may have a different file for each project. Create a projname.wdw file and put in the same directory as your project file (.WDP). To use the same file for all projects, create or modify the DEFAULT.WDW file in the USER folder. In the Project Manager, right-click the project name and select Settings to find the full path.

 

There should be a separate line in the file for each AutoCAD Electrical wire layer. The line has three fields, each field separated by a semi-colon. The first field is the actual wire layer name used on the drawing. The second field is the wire layer description. This description is used in the AutoCAD Electrical Wire Colour/Gauge Label tool. The third field is the wire size.

 

For example, if you have a wire layer called 14_RED_THHN and you want the wire colour/gauge label to read #14AWG RED for this layer, and the wire itself has a wire diameter of 0.0087, the line in the .WDW file would read:

 

14_RED_THHN;#14AWGRED;0.0087

 

<HELP FILE END>

 

Going to a conversion site that gives a list AWG, Imperial and Metric wire sizes gives an imperial size of 0.0641 for #14AWG

 

https://technick.net/guides/electronics/awg_to_metric/

 

Therefore the lines in your WDW file for #14AWG wires should read

 

WHT_14_THHN;#14AWG WHT;0.0641

RED_14_THHN;#14AWG RED;0.0641

BLK_14_THHN;#14AWG BLK;0.0641

YEL_14_THHN;#14AWG YEL;0.0641

 

This should then give you a correct % Full reading for the conduit size.

 

Regards Brad

 

 

>

Brad Coleman, Electrical Draftsman
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Message 3 of 12

galashkina
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Thanks, Brad!

I read Help file before asking a question.

But I did not understood by a formula calculated % Full.

I corrected the WDW - file according to your recommendations.

 

1.png

 

I added a line to the ww1 - file

 

2.png

 

Now the Conduit is filled to 5%. This is an incorrect value.

 

3.png

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Message 4 of 12

Icemanau
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The original values for the WW1 file should be correct as they are dealing with conduit sizes and not wire sizes.

 

Try this spreadsheet which deals with this and actually gives some size values for schedule 40 PVC conduit.

 

http://www.google.com.au/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&cad=rja&uact=8&ved=0ahUKEwiqjs6-lc...

 

Also, I don't believe that they make a 1/4' conduit. It would be to small to actually be useful.

The smallest I have seen is the 16mm or 11/2' and for me, the 16mm is actually very rarely used.

An electrician would normally use 20mm as the smallest unless there is a specific reason to use the 16mm.

 

Most wiring regulations call for a minimum percentage of the conduit to be free space.

1/4' conduit would have a usable are of around that of a pencil.

This would not be enough space for those regulations regarding free space in the conduit

 

 

Regards Brad

>

Brad Coleman, Electrical Draftsman
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Message 5 of 12

galashkina
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I got the following:

 

File WDW

 

1.png

 

File WW1

 

2.png

 

Inner box size 1 inch: 0.34 (inches) х 25.4 х 0.8 = 6.9088 мм

 

The size of 15 wires YEL_14_THHN: 0.0641 х 15 = 0.9615

 

% Full: 0.9615 / 6.9088 х 100 = 13.9170

 

3.png

 

My result is consistent with the calculation program.

 

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Message 6 of 12

galashkina
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this is from the Help file...

"For example, if you have a 1-inch conduit with an inner diameter of 0.8...."

 

My calculation:

Inner box size 1 inch: 0.34 (inches) х 25.4 х 0.8 = 6.9088 мм

 

 

 

 

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Message 7 of 12

galashkina
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Still, I doubt the correctness of my calculation Smiley Sad

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Message 8 of 12

galashkina
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What nonsense I have written above! Smiley Embarassed

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Message 9 of 12

rhesusminus
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Accepted solution

Hi Galashkina.

 

Unfortunately, the values needed for the conduit size and wire size are two different "units"

.

 

In the WW1, you specify the AREA inside the conduit. For a 1" conduit, with 0.8" inside DIAMETER (Radius = 0.4"), the value to specify in the wdw-file is:

 

πr² = 3.1415 * (0.4 * 0.4) = 0.5026 square inches

 

This give this line in the WW1-file:

1";0.5026

 

 

In the WDW you specify the DIAMETER of the wire (remember to include the diameter of the wire, including the insulation).

So, a 14 AWG is about 0.0641" in diameter. lets add insulation , 0.0236" and we get a total diameter of 0.1113" (0.0236"+0.0236"+0.0641")

 

This gives this line in the WDW-file:

YEL_14_THHN;#14AWG YEL;0.1113

 

1 piece of 14 AWG wire will now fill the 1" conduit about 2-3%, because the AREA of the wire is 0.0097 square inches.

(I'm not sure how many decimals they operate with in PI, and it seems like the round up to the nearest integer)

 

 

 

 

 

 

 


Trond Hasse Lie
EPLAN Expert and ex-AutoCAD Electrical user.
Autodesk Expert Elite Alumni
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Message 10 of 12

galashkina
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Thank you, rhesusminus !

 

Your calculation is correct.

 

2.png

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Message 11 of 12

Anonymous
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Is this table you show for conduit size and % fill in autocad electrical.

What type of schedule can you create from this data input? Does it allow you to create a simple conduits schedules with wire and percent fill?  I am starting a project and need a

software that can create these things.

 

Thank you

 

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Message 12 of 12

galashkina
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If you correctly prepare the data for the calculation, it will work.
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