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

4

V=

760

= 190.0 feet per minute (fpm)

4

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)

of a circle: A = πr2, where π = 3.1416 and r = radius of the circle (r = 1/2 x diameter).

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

MD0165

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