Q = AV
Where Q = rate of flow in cfm.
V = average linear velocity in ft/min.
A = cross-sectional area of the duct or hood in ft2.
From this equation, it is possible to calculate airflow rate (Q) if the velocity (V) and the
cross-sectional area (A) are known.
For example: If a duct is 12" by 36," and the velocity of the air has been measured four
times, with readings of 170, 185, 210, and 195 ft/min, what is the rate of flow?
First, determine the average velocity:
V = 170 + 185 + 210 + 195
= 190.0 feet per minute (fpm)
Then, determine cross-sectional area:
A = 1' x 3' = 3 ft2 (Change inches to feet before computing A.)
Finally, substitute values in the formula:
Q = AV
Q = 3 x 190.0 = 570.0
Q = 570.0 cubic feet per minute (cfm)
NOTE: For circular ducts, cross-sectional area is determined using the formula for area
of a circle: A = πr2, where π = 3.1416 and r = radius of the circle (r = 1/2 x diameter).
c. Air Velocity. The purpose of the exhaust system in an industrial work area is
to remove the contaminant from the environment. To do this effectively, the air must
have sufficient velocity to overcome opposing air currents and particle inertia causing
the contaminated air to flow into the enclosure (usually a hood) and there must be
sufficient velocity to keep the contaminant from settling out in the duct.
(1) Capture velocity. To effectively remove the contaminant from the
environment, the air must have sufficient velocity to overcome opposing air currents and
particle inertia causing the contaminated air to flow into the enclosure (usually a hood).
This minimum air velocity is called the capture velocity and is an important
consideration in designing a local exhaust ventilation system or in evaluating the
operating effectiveness of an existing system. The minimum air velocity at a point
within or in front of an exhaust hood necessary to overcome opposing air currents and