Conduit Fill Calculator
Check conductor fill percentage in a conduit.
About the Conduit Fill Calculator
Calculates the percentage fill of a conduit given the conduit's internal cross-sectional area and the number and size of conductors, and flags the NEC fill limits.
$$ fill = \frac{n \cdot a_{wire}}{a_{conduit}}\times 100 $$
How to use this calculator
- Enter the number of conductors, then enter the cross-sectional area of one wire as a square area value.
- Enter the conduit’s inside cross-sectional area using the same units.
- Calculate the fill percentage with \(fill = \frac{n \cdot a_{wire}}{a_{conduit}} \times 100\).
- Compare the result to the allowed fill limit for your application before installing.
The formula explained
The formula computes the percentage of the conduit area that is taken up by all the wires combined. It compares total wire area, \(n \cdot a_{wire}\), to the conduit area, \(a_{conduit}\).
- n = the number of conductors in the conduit
- \(a_{wire}\) = the cross-sectional area of one wire
- \(a_{conduit}\) = the inside cross-sectional area of the conduit
- fill = the conduit fill percentage
Step by step method
- Find the area of one wire and multiply it by \(n\) to get the total wire area.
- Divide the total wire area by \(a_{conduit}\).
- Multiply by \(100\) to convert the result to a percentage.
Worked example
Problem. A conduit contains \(4\) wires, each with an area of \(0.12\,\text{in}^2\). The conduit’s inside area is \(1.20\,\text{in}^2\). Find the fill percentage.
- Compute the total wire area: \(4 \cdot 0.12 = 0.48\,\text{in}^2\).
- Divide by the conduit area: \(\frac{0.48}{1.20} = 0.4\).
- Convert to a percent: \(0.4 \times 100 = 40\%\).
Answer. \(40\%\)
Tips and common mistakes
- Make sure all areas use the same units, such as \(\text{in}^2\) or \(\text{mm}^2\).
- Do not confuse the outside conduit size with the inside area, because fill uses the inside cross-sectional area.
Frequently asked questions
What are the NEC limits?+
≤53% for one conductor, ≤31% for two, ≤40% for three or more.
Why limit fill?+
To allow heat dissipation and let conductors be pulled without damage.
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