Raceway fill Examples 2 and 3

Example 2

What size conduit is required for 10 cables and 20 Number 16 PTF Conductors installed in Rigid Metal Conduit?

Cable Major Diameter =.250 inches
Cable Minor Diameter =.200 inches
20 PTF NO. 16 CU

Answer

Step 1: Find the total cross sectional area of all the conductors.

Area of # 16 PTF is .0075 X 20 conductors .150 square inches.
 

Step 2: Find the total cross sectional area of all the cables.

Cross Sectional Area = (3.14)(.125)(.125) CSA =.0491 square inches
CSA for 10 cables = (10) (.0491)
CSA for 10 cables =.491 square inches
 

Step 3: Find the total cross sectional area of all the cables and wires.

Total CSA =.150 +.491 square inches.  Total CSA = .641 square inches.
 

Step 4: Find the size of Raceway.

Turn to Table 4, "Dimension and Percent Area of Conduit and of Tubing," in the Section titled "Rigid Metal Conduit." The question addresses over 2 wires so read down the column heading "Over 2 Wires 40% Sq.  In." Read down this column until you find the number that is greater than .641. After finding  .829, read across (left) to the "Trade Size Inches" column that indicates that a One and one half inch conduit is required.

Example 3

How many cables with a .200 inch diameter can be installed in a 2 inch PVC Conduit Schedule 80 nipple 18 inches long?
 
 

Step 1:

Find the 60 per cent fill cross sectional area of the PVC conduit by using table 4 in Chapter 9.

From the Table, the 100 per cent cross sectional area is 2.874 square inches.  Use 60 per cent rill since this is allowed for nipples no longer than 24 inches per note 4 to Table I of Chapter 9.

.6 x 2.874 = 1.7244 square inches
 

Step 2:

Find the cross sectional area of one cable.

CSA = (3.14) (.100)(.100)
CSA =.0314 square inches
 

Step 3:

Find the number of cables that can be installed.

Number of Cables = 1.7244/.0314 1.7244/.0314 = 54.92

By note 7 to Table 1 of Chapter 9 round up if division yields a decimal of .8 or larger.

Therefore 55 cables can be installed in the 2 inch nipple.