Separate duct airflow area from duct wrapping area

A duct's cross-sectional area helps relate airflow to velocity. Its outside surface area helps estimate wrapping. Using one area in place of the other can produce a result with the right-looking units but the wrong meaning.

Calculate a round duct example

For an illustrative 0.2 m internal diameter, cross-sectional area is π × 0.1² = 0.031416 m². At an assumed 0.1 m³/s airflow, mean velocity is approximately 3.18 m/s. This is a continuity calculation, not a recommended duct velocity.

Measure the outside for wrap

A straight round duct with 0.2 m outside diameter and 5 m length has approximately π × 0.2 × 5 = 3.142 m² of bare outside surface. Insulation thickness, overlaps and fittings can increase the wrap requirement. Use outside dimensions for the surface calculation.

Handle rectangular duct separately

For a straight rectangular duct 0.4 m wide, 0.25 m high and 5 m long, the four-side surface is 2 × (0.4 + 0.25) × 5 = 6.5 m² before ends, fittings and laps. Elbows and transitions need separate takeoffs.

Keep design decisions outside the quantity

The calculations do not select fan pressure, duct dimensions, insulation thickness, vapour control or fire performance. Obtain those details from the system specification. Record whether each diameter is internal or external so the airflow and wrapping checks can be independently repeated.

Found a measurement or calculation issue? Send the example and your correction.