VENTILATION REQUIREMENTS
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4
IO-VENT-SS
Determine Your
Ventilation Needs
Good indoor air quality is based
in part on the capacity of the
home’s ventilation system.
The two most common
methods for determining the
ventilation needs of a home are
the Room Count Calculation
method and the Air Change
Per Hour method. Both
methods calculate an
approximate ventilation rate
(in CFM or L/s). Their calculation
methods are described on
this page.
A. Room Count Calculation Method
LIVING SPACE
NUMBER OF ROOMS
CFM (L/S)
CFM REQUIRED
Master Bedroom
————
x 20 cfm (10 L/s)
=
————
Basement
————
x 20 cfm (10 L/s)
=
————
Single Bedroom
————
x 10 cfm (5 L/s)
=
————
Living Room
————
x 10 cfm (5 L/s)
=
————
Dining Room
————
x 10 cfm (5 L/s)
=
————
Family Room
————
x 10 cfm (5 L/s)
=
————
Recreation Room
————
x 10 cfm (5 L/s)
=
————
Other
————
x 10 cfm (5 L/s)
=
————
Kitchen
————
x 10 cfm (5 L/s)
=
————
Bathroom
————
x 10 cfm (5 L/s)
=
————
Laundry Room
————
x 10 cfm (5 L/s)
=
————
Utility Room
————
x 10 cfm (5 L/s)
=
————
TOTAL ventilation requirement (add last column) =
1 CFM = 0.47189 L/s
1 L/s = 3.6 m3/hr
B. Air Change Per Hour Calculation Method
TOTAL volume of the house (in cubic feet) X 0.35 air changes/hour = CUBIC FT/HR
Take total CUBIC FT/HR and divide by 60 to get CFM
Example:
A 25'x 40' house with basement:
25' X 40' = 1,000 sq. ft.
1,000 sq. ft. x 8' room height x 2 storeys
(1st floor + basement) = 16,000 cu. ft.
16,000 cu. ft. x 0.35 ACH = 5,600 cu. ft.
5,600 cu. ft. / 60 = 93.3 CFM
93.3 CFM is your ventilation requirement
