You can think of the air and contaminants flowing through ductwork as if it were like a car moving down a highway. Expressways don’t have sharp 90 degree turns or abrupt changes in width. These would cause a slow down in traffic, unless of course an accident is in the way. Expressways also tend to be rather large thoroughfares. The science behind ductwork follows the same basic principles to work effectively. It will minimize or eliminate sharp turns and it will avoid any abrupt changes in diameter. It will also be as wide in diameter as the environment will accommodate in order to move air volume most effectively.
A local exhaust ventilation system’s performance can be greatly hampered and workers exposed to hazards if ductwork leaks. And if the leaks are upstream of the fan and large enough, they can reduce the ability of the local exhaust ventilation system to draw the airborne contaminants into the hood. Air starts getting drawn in through the leaks rather than through the hood. If the leaks are downstream of the fan, the airborne contaminants can re-enter the work area through the leaks rather than going outside through the exhaust stack.
Ducts come in an endless variety of diameters, the diameter being part of a simple mathematical equation relating to flow velocity. In the simplest of terms, the flow of air through ductwork is governed by the following equation:
Q = V × A
where Q is the flow rate of air through the duct in cubic feet per minute (CFM), V is velocity of the air flow in feet per minute, and A is the cross sectional area of the duct in square feet.As an example, suppose you want to design a local exhaust ventilation system with ducts no greater than 5 inches in diameter because of space and clearance limitations. You want to use round ducts for the system because they handle air more efficiently and have no sharp corners where dust can collect. If an industrial hygienist determines that the air is required to flow at a minimum of 800 feet per minute through the duct, what is the airflow rate through the duct? Well, since we are dealing with a round duct, its cross sectional area is that of a circle:
A = (π × d2) ÷ 4
where d is the diameter of the inside of the duct as shown in Figure 1 below.
Figure 1 – Cross Section of a Round Duct
So to use this equation for area, to solve for Q, then we must first convert the duct diameter from inches to feet, which makes our equation look like this:
5 inches ÷ 12 inches per foot = 0.416 feet
This gives us a duct cross sectional area of:
A = (π × (0.416 feet) 2) ÷ 4 = 0.136 square feet
And the air will flow through the duct at this rate:
Q = V × A = 800 ft./minute × 0.136 ft.2 = 108.8 CFM
This air flow rate is good to know, because it will help the
designer to select an appropriate fan for the local exhaust ventilation
system. This is because fans are listed in manufacturers’ catalogs
according to how many CFM they can handle.Next time, we’ll learn more about the rest of the local exhaust ventilation system, namely, the filter, fan, and exhaust stack.
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