Box Fill Calculator

Enter the required parameters to precisely calculate “Box Fill" requirements for an electrical wiring box.

Home > Everyday-Life > Box Fill Calculator

Largest conducting wire size

Are you using a box with internal clamps

Number of conducting wires

wires

Will you need some support fittings

Number of devices

devices

Number of grounding conductors

wires

Largest grounding wire size

what is box Fill Calculation?

it's miles the calculation of the combined extent occupied via undertaking wires, grounding wires, and other components in an electrical software box.

The main objective of this field fill calculation is to prevent overcrowding in electrical boxes due to the fact an overloaded container can bring about faults, arcing, or maybe pose a chance of hearth.

Calculate field Fill:

placed those values in the formulation to calculate the fill extent of each thing and sum all of the fill volumes to locate the proper field size

Now, permit's calculate the extent allowances for every factor within an electrical box

Conductor Fill:

$\ A_w =\ n_w$

And

$\ V_w =\ A_w\times \ V_{largest\ conductor}$

Where

$A_w$ is the conductor fill volume allowance
$\ n_w$ represents the wide variety of accomplishing wires
$V_w$ is the conductor fill volume
$V_{largest\ conductor}$ is the loose space we offer for the biggest conductor

in the following desk we've got furnished the conductor wire size and their allowance volumes. For ease, use our wire length calculator and optimize your wire size assessments.

Size of Conductor (AWG)	Volume Allowance Required Per Conductor (in.³)
18	1.5
16	1.75
14	2
12	2.25
10	2.5
8	3
6	5

Clamp Fill:

if you are the usage of an electrical container with inner clamps, then:

$\ A_c=\ 1$

Otherwise (no clamps):

$\ A_c=\ 0$ $\ V_c =\ A_c\times \ V_{largest\ conductor}$

Where

$\ A_c$ is the clamp fill volume allowance
$\ V_c$ is the clamp fill volume

support Fittings Fill:

If there are one or more luminaire studs or hickey within the box:

$\ A_s=\ 1$ If

There are no luminaire studs or hickey in the field:

$\ A_s=\ 0$ $\ V_s =\ A_s\times \ V_{largest\ conductor}$

Where

$\ A_s$ represents the assist fittings fill volume allowance
$\ V_s$ is the guide fittings quantity

device or gadget Fill:

$\ A_d = 2 * n_d$

Where

$\ A_d$ is the representative of the device fill quantity allowance
$\ n_d$ is the wide variety of devices

$\ V_d =\ A_d\times \ V_{largest\ conductor}$

Where

$\ V_d$ is the device fill volume

gadget Grounding Conductor Fill:

If 4 grounding conductors that are entering in the electrical field,

$\ A_g=\ 1$

If 5 or greater:

$\ A_g=\ 1+\dfrac{n_g−4}{4}$

$\ A_g=\dfrac{n_g}{4}$

Where

$\ A_g$ is the consultant of the grounding conductor fill volume allowance
$\ N_g$ is the entire wide variety of grounding conductors

In equation form, it could be expressed as:

$\ V_g =\ A_g\times \ V_{largest\ ground\ Wire}$

Where

$\ V_g$ indicates the device grounding conductor fill volume
$\ A_g$ represents the grounding conductor fill extent allowance
$\ V_{largest\ ground\ Wire}$ is the loose area

the whole field Fill volume:

$\ V_{total} =\ V_w+V_c+V_s+V_d+V_g$

If the most important conductor is identical to the most important grounding cord size then discover the entire extent allowance and multiply it by the loose space required for the most important conductor.

$\ A_{total} =\ A_w+A_c+A_s+A_d+A_g$

$\ V_{total} =\ A_{total}\times \ V_{largest\ conductor}$

Where

$\ V_{total}$ is the total container fill
$\ A_{total}$ represents the entire quantity allowance
$\ V_{largest\ ground\ Wire}$ is the unfastened area as described in table 314.16(B)

Example:

suppose you have a field with 10 wires, wherein the largest conducting wire size is 12 AWG, the variety of devices is 5, and there are 10 grounding conductors with the most important grounding twine size being 12 AWG. You want to decide the subsequent:

Grounding wire volume
Equipment grounding fill volume
Total volume allowance needed
Total box fill volume

Solution:

Given:

Largest conducting wire size = 12 AWG = 2.25 cu inches
Number of conducting wires = 10
Devices = 5
Grounding conductors = 10
Largest Grounding conductor wire size = 12 AWG = 2.25 cu inches

Conductor Fill Volume:

For conducting wires, the fill volume is calculated by:

$V_w = A_w \times V_{\text{largest conductor}}$

$A_w = 10$ (number of conducting wires)

$V_w = 10 \times 2.25 = 22.5$ cu inches

Device Fill Volume Allowance:

The device fill volume allowance is:

$A_d = 2 \times n_d$

$A_d = 2 \times 5 = 10$

Device Fill Volume:

For the devices:

$V_d = A_d \times V_{\text{largest conductor}}$

$V_d = 10 \times 2.25 = 22.5$ cu inches

Equipment Grounding Fill Volume Allowance:

The equipment grounding fill volume allowance is:

$A_g = \dfrac{n_g}{4}$

$A_g = \dfrac{10}{4} = 2.5$

Grounding Wire Volume:

The grounding wire volume is calculated as:

$V_g = A_g \times V_{\text{largest ground wire}}$

$V_g = 2.5 \times 2.25 = 5.625$ cu inches

Total Volume Allowance Needed:

The total volume allowance needed is:

$A_{\text{total}} = A_w + A_d + A_g$

$A_{\text{total}} = 10 + 10 + 2.5 = 22.5$ cu inches

Total Box Fill Volume:

The total box fill volume is:

$V_{\text{total}} = V_w + V_d + V_g$

$V_{\text{total}} = 22.5 + 22.5 + 5.625 = 50.625$ cu inches

Ensure precision by cross-checking your manual calculations with our electrical box fill calculator and optimize your results for a safe electrical system.

FAQ’s:

Do Grounds depend In container Fill?

sure, the grounding wires are also covered in calculating container fill volumes. in line with the NFPA 70: national electrical Code® 2020, a unmarried volume allowance is particular for one to 4 equipment grounding conductors or device leaping conductors. each extra grounding conductor wishes an additional 1/four volume allowance.

what's the NEC Code For box Fill?

The box fill code, outlined in NEC 314.sixteen states that the containers and conduit our bodies ought to have the appropriate accredited size to offer loose space to all the enclosed conductors. The quantity of the field calculated as per 314.sixteen(A) need to never be less than the fill calculation as determined in 314.sixteen(B) in any case.