 NEC 314.16 Part (A): Box Volume Calculations,
 NEC 314.16 Part (B): Box Fill Calculations,
 NEC 314.16 Part (C): Conduit Bodies.
Today, I will explain Conduit Fill Calculations as follows.
You can review the following articles in the same course for more information:
 Types of Electrical Conduits
 Conduit Fittings and Supports
 Electrical Boxes  Part One
 Electrical Boxes – Part Two
Conduit Fill Calculations
Conduit Sizes
Designations
The
conduits have two size designations as follows:
Table
300.1(C) identifies a distinct metric designator for each circular raceway
trade size.

Tables
used for Conduit Fill Calculations
First:
Chapter 9 which
includes the following tables:
You can
download a PDF copy of Chapter 9 Tables by click on the link.
Second:
Annex C Tables
Informative
Annex C contains conductor fill tables for each of 12 types of conduit and
tubing. The Informative Annex C tables — which are based on the dimensions
given in Tables 1 and 4 of Chapter 9 for conduit and tubing fill and on the
dimensions for conductors in Table 5 of Chapter 9 — provide conductor fill
information based on the specific conduit or tubing and on the conductor
insulation type, size, and stranding characteristics. Examples of how to use
these tables are included in the commentary both here and in Informative
Annex C.
You can
download a PDF copy of Annex C Tables by click on the link.

Chapter 9  Table 1
Table 1
establishes the maximum fill permitted for the circular conduit and tubing
types. It is the basis for Table 4 and for the information on conduit and
tubing fill provided in the Informative Annex C tables.
Informational
Note No. 1:
The
installation of conductors in a conduit can face some difficulties due to:
So, it is
recommended that where a difficult installation is anticipated due to above
reasons, the available solutions will be as follows:
Informational
Note No. 2:
Jam ratio = ID of raceway / OD of conductor
To avoid
difficult conductor installations and potential conductor insulation damage
due to jamming within the conduit or tubing, a jam ratio between 2.8 and 3.2
should be avoided.
As an
example:
Table C.1 in Informative Annex C permits three 8 AWG conductors in
trade size 1⁄2 electrical metallic tubing (EMT). An 8 AWG conductor has an
outside diameter (OD) of 0.216 in. (from Table 5), and a 1⁄2 in. EMT has an
internal diameter (ID) of 0.622 in. (from Table 4).
The jam
ratio is calculated as follows:
Jam ratio
= ID of raceway / OD of conductor = 0.622 / 0.216 = 2.88
So,
Jamming of conductors will occur, use the next larger trade size conduit.
A 3⁄4 in.
EMT has an internal diameter (ID) of 0.824 in. (from Table 4).
So, Jam
ratio = ID of raceway / OD of conductor = 0.824 / 0.216 = 3.815

Chapter 9
 Table 4

Notes to
chapter 9 Tables
(1) See
Informative Annex C for the maximum number of conductors and fixture wires,
all of the same size (total crosssectional area including insulation)
permitted in trade sizes of the applicable conduit or tubing.
(2) Table
1 applies only to complete conduit or tubing systems and is not intended to
apply to sections of conduit or tubing used to protect exposed wiring from
physical damage.
(3)
Equipment grounding or bonding conductors, where installed, shall be included
when calculating conduit or tubing fill. The actual dimensions of the
equipment grounding or bonding conductor (insulated or bare) shall be used in
the calculation. The dimensions of bare conductors are given in Table 8.
(4) Where
conduit or tubing nipples having a maximum length not to exceed 600 mm (24
in.) are installed between boxes, cabinets, and similar enclosures, the nipples
shall be permitted to be filled to 60 percent of their total crosssectional
area, and 310.15(B)(3)(a) adjustment factors need not apply to this
condition.
(5) For
conductors not included in Chapter 9, such as multiconductor cables, high
voltage Cables and optical fiber cables, the actual dimensions shall be used.
The
crosssectional area can be calculated in the following manner, using the
actual dimensions of each conductor:
Crosssectional
area = d
2
cmil
Where:
d = outside
diameter of a conductor (including insulation)
1 in. =
1000 mil
1 cmil
(circular mil) = Ï€/4
(3.1416/4) square mil =0.7854 square mil.
Conversion
from square millimeters to circular mils:
To convert
from square millimeters to circular mils (approximately) follows:
k = 1973.53
circular mils / mm^{2}
(6) For
combinations of conductors of different sizes, use Table 5 and Table 5A for
dimensions of conductors and Table 4 for the applicable conduit or tubing
dimensions.
(7) When
calculating the maximum number of conductors permitted in a conduit or
tubing, all of the same size (total crosssectional area including
insulation), the next higher whole number shall be used to determine the
maximum number of conductors permitted when the calculation results in a
decimal of 0.8 or larger.
(8) Where
bare conductors are permitted by other sections of this Code, the
dimensions for bare conductors in Table 8 shall be permitted.
(9) A
multiconductor cable or flexible cord of two or more conductors shall be
treated as a single conductor for calculating percentage conduit fill area.
For cables that have elliptical cross sections, the crosssectional area
calculation shall be based on using the major diameter of the ellipse as a
circle diameter.

Example#1:
Three 15kV single conductors are to be installed in rigid metal conduit (RMC). The outside diameter of each conductor measures 15⁄8 in., or 1.625 in. What size RMC will accommodate the three conductors?
Solution:
Step 1: Find the crosssectional area within the conduit to be displaced by the three conductors:
1.625 in. x 1.625 in. x 0.7854 x 3 = 6.2218 in.2 or 6.222 in.2
Step 2: Determine the correct conduit size to accommodate the three conductors. Table 1 allows 40 percent conduit fill for three or more conductors, and Table 4 indicates that 40 percent of trade size 5 RMC is 8.085 in.2.
Thus, trade size 5 RMC will accommodate three 15kV single conductors.
Example#2:
What traditional wire size does the size 125 mm2 represent (approximately)?
Solution:
Circular mil area = wire size (mm2) x conversion factor = 125 mm2 x 1973.53 circular mils / mm2 = 246,691 circular mils or 246.691 kcmil
Therefore, the 125 mm2 wire is larger than 4/0 AWG (211.6 kcmil) but smaller than a 250kcmil conductor.
Notes to example#2:
 If a 125 mm2 wire is determined to be the minimum or recommended size conductor, it is important to understand that size 250 kcmil would be the only Table 8 conductor with equivalent crosssectional area because 4/0 AWG is simply not enough metal.
 It is important, however, to note that the 250kcmil conductor ampacity could not be used for a 125 mm2 conductor, because the metric conductor size is smaller. The 4/0 AWG ampacity can be used, or the ampacity can be calculated under engineering supervision.
Example#3:
A 200ampere feeder is routed in various wiring methods [electrical metallic tubing (EMT); rigid polyvinyl chloride conduit (PVC), Schedule 40; and rigid metal conduit (RMC)] from the main switchboard in one building to a distribution panelboard in another building. The circuit consists of four 4/0 AWG XHHW copper conductors and one 6 AWG XHHW copper conductor. Select the proper trade size for the various types of conduit and tubing to be used for the feeder.
Solution:
The used tables are:
 Table 1
 Table 1, Note 6 refers to Table 5 for the area required for each insulated conductor.
 Note 6 also refers to Table 4 for selection of the appropriate trade size conduit or tubing.
 Table 4 contains the allowable cross sectional area for conduit and tubing based on conductor occupied space (40 percent maximum in this example).
Step 1: assign the fill percentage from table 1
All the raceways for this example require conduit fill to be calculated according to Table 1 in Chapter 9, which chapter 9 table 1 permits conduit fill to a maximum of 40 percent where more than two conductors are installed.
Step 2: Calculate the total area occupied by the conductors, using the approximate areas listed in Table 5:Four 4/0 AWG XHHW: 4 x 0.3197 in.2 = 1.2788 in.2
One 6 AWG XHHW: 1 x 0.0590 in.2 = 0.0590 in.2Total area = 1.3378 in.2 or 1.338 in.2
Step 3: Determine the proper trade size EMT, RMC, and PVC (Schedule 40) from Table 4.
The portion of this feeder installed in EMT requires a minimum trade size 2, which has 1.342 in.2 of available space for over two conductors. The minimum required space is 1.338 in.2, which is less than the trade size 2 EMT 40 percent fill.
RMC also requires a minimum trade size 2, because trade size 2 RMC has 1.363 in.2 of available space for over two conductors. PVC (Schedule 40), however, requires a minimum trade size 2 1⁄2.
Trade size 2 PVC has 1.316 in.2 allowable space for over two conductors and is less than the 1.338 in.2 required for this combination of conductors. Therefore, it is necessary to increase the PVC size to 2 1⁄2 trade size, the next standard size increment.
Example#4:
Determine how many 10 AWG THHN conductors are permitted in a trade size 1 1⁄4 rigid metal conduit (RMC).
Solution:
Table 1 permits 40 percent fill for over two conductors.
From Table 4, 40 percent fill for trade size 11⁄4 RMC is 0.610 in., and from Table 5, the crosssectional area of a 10 AWG THHN conductor is 0.0211 in.2. The number of conductors permitted is calculated as follows:
0.610 in.2 / 0.0211 in.2 per conductor = 28.910 conductors
Based on the maximum allowable fill, the number of 10 AWG THHN conductors in trade size 1 1⁄4 RMC cannot exceed 28. However, in accordance with Note 7, an increase to the next whole number of 29 conductors is permitted in this case, because 0.910 is greater than 0.8.In this case the number of conductors permitted = 29 conductors
Note to example#4:
Although increasing the total to 29 conductors results in the raceway fill exceeding 40 percent, the amount by which it is exceeded is a fraction of 1 percent and will not adversely affect the installation of the conductors.