1. Mathematics of Air Filtration
Presented by Rick Peckham
2010 NAFA Annual Convention
Hawaii – August 8th, 2010

2. Airflow Quantity and Velocity
Quantity of Air “Q” (expressed as CFM)
Velocity of Air “V” (expressed as FPM)
Face Area “A” (expressed as Sq. Ft.)
With 2 known values – 3rd can be calculated

3. Calculating Quantity (CFM)
When face area and velocity are known the
CFM can be determined.
4 x 500 = 2000
4 sq. ft. x 500 FPM = CFM

4. Calculating Velocity (FPM)
When face area and air quantity are known
the velocity can be determined.
2000 / 4 = 500
2000 CFM / 4 sq. ft. = 500 FPM

5. Calculating Face Area (sq. ft.)
When the velocity and quantity of air
are known, the face area can be
determined.
2000 / 500 = 4
2000 CFM / 500 FPM = 4 sq. ft.

6. Application Example 40,000 CFM / 400 fpm = 100 sq. ft.
An engineer wants to know how many filters
he needs to use for a system that delivers
40,000 CFM - desired velocity is 400 fpm.
Calculations
40,000 CFM / 400 fpm = 100 sq. ft.
Assume a 24x24 face size with 4 sq. ft.
100 sq. ft / 4 sq. ft. = 25 filters

7. Another Method Knowing he wants 400 fpm you would take
the face area of a 24x24 which is 4 and
multiply x 400 which equals 1600 CFM/filter
40,000 total system CFM divided by 1,600
CFM per filter also equals 25 filters.

8. Square Inches to Square Feet
When dealing with a 24x24 it is easy
to come up with the face area (2x2=4)
Other sizes require conversion
Divide square inches by 144 (sq in/sq ft)
20x20 = 400 / 144 = round to 2.78
20x25 = 500 / 144 = round to 3.47

9. Application Example An engineer doesn’t have CFM data for his
system and wants to know what it is. He has
measured the velocity at 375 fpm and the
unit contains x20 filters.
Calculations
Convert sq. in. to sq. ft. 20x20 = 400 / 144
which equals round to 2.78 sq. ft.
24 x 2.78 = total sq. ft.
66.72 sq. ft. x 375 fpm = 25,020 CFM

10. Application Example A maintenance lead is having air delivery
problems and wants to lower the pressure
drop across his filter bank. He has room to
expand the filter bank. The bank contains
10 filters that were originally MERV 7 pleats
but have been upgraded to MERV 13. System
CFM is 24,000 and he wants to reduce his
velocity to 300 fpm to cause a big reduction in
pressure drop. How many filters does he need?

11. Calculation Examples Alternate Method
He is asking for 300 fpm velocity
4 x 300 = 1200 CFM per 24x24 filter
Total system CFM 24,000/1,200 = 20 filters
Alternate Method
24,000 / 300 = 80 sq. ft. face area needed
80 / 4 = 20 filters needed.

12. Media Velocity Media velocity is calculated by starting with
the CFM per filter and dividing by the total
media area. It is an important consideration
because lower media velocities normally
improve the efficiency of a filter. Excessive media
velocities can adversely affect media performance!
Pleat – 2,000 CFM / 17.6 sq. ft. = FPM
113.6 is the media velocity for this pleat
Bag or pocket filter – 2,000 CFM / 90 sq. ft.
equals FPM media velocity

13. Additional Examples Box – 2,000 CFM / 58 sq. ft. = 34.48 FPM
Cell – 2,000 CFM / 100 sq. ft. = 20 FPM
4V – 2,000 CFM / 194 sq. ft. = FPM
Hepa – 1,000 CFM / 160 sq. ft. = 6.25 FPM

14. Practical Use Some specifications identify media velocity
Reverse engineering – if you know the
correct range of velocity for a specific media
you can calculate the media area needed
for desired performance characteristics at
various airflows.

15. Application Example Customer wants a special filter with 90% or
MERV 14 fiberglass media. Size is to be
20x25x2 and he wants to know how much
CFM the filter will handle at .50 IPD.
Assume a MERV 14 glass bag with 90 sq. ft.
of media has an IPD of 2000 CFM

16. Application Example Calculate media velocity
2,000 CFM / 90 sq. ft. = FPM
The 20x25x2 filter will contain 10 sq. ft. of
media area.
10 sq. ft. x fpm = CFM

17. Air Changes Per Hour (ACH)
This calculation is used extensively in the
Cleanroom industry. It impacts the overall
air quality since the more times the air is
changed the more potential there is for the
reduction of contaminants in the conditioned
space.

18. Air Changes Per Hour (ACH) For Existing Room
Information needed
Volume of conditioned space
Length x Width x Height = Cubic Feet
CFM being delivered converted to CFH
2,000 CFM x 60 minutes/hour = 120,000 CFH
Divide CFH by volume in Cubic Feet = ACH

19. Application Example Existing Room
Room is 25 ft. x 50 ft. x 8 ft.
At 2,000 CFM what are the air changes/hour
25 x 50 x 8 = 10,000 cubic feet
2,000 CFM x 60 min/hr = 120,000 CFH
120,000 CFH / 10,000 cubic feet = 12 ACH

20. Application Example Existing Room
Again, the room is 25 ft. x 50 ft. x 8 ft.
and the total amount of air delivered is 2,000 CFM
Think of it this way – 2,000 CFM fills the
room completely every 5 minutes
10,000 cubic feet / 2,000 CFM = 5 minutes to fill
60 minutes per hour divided by 5 minutes to fill
also equals 12 ACH – the room gets filled every
5 minutes so the air is changed 12 times per hour.

21. Application Example Existing Room
A homeowner wants to know how many
ACH he is getting in his house. House is
80 x 30 x 8 = 19,200 cubic feet of space.
He has a 5 ton unit delivering 2,000 CFM.
2,000 CFM x 60 = 120,000 CFH
120,000 / 19,200 = 6.25 ACH

22. Same Application Example Alternate Calculation
A homeowner wants to know how many
ACH he is getting in his house. House is
80 x 30 x 8 = 19,200 cubic feet of space.
He has a 5 ton unit delivering 2,000 CFM.
19,200 cubic feet / 2,000 CFM = 9.6 min
The house fills with air every 9.6 minutes
60 min/hr divided by 9.6 min = 6.25 ACH

23. Air Changes per Hour Room Design
Information needed
Volume of design space
Length x Width x Height = Cubic Feet
Desired ACH x Cu. Ft. = Total volume (CFH)
Total Air Volume in CFH converted to CFM
Divide Total Air Volume by 60 =CFM needed

24. Application Example Room Design
An HVAC engineer wants 25 ACH in the
room he is designing. The room dimensions
are 50 ft. x 100 ft. x 12 ft. What amount of
CFM will he need to provide?
Find the cubic volume
50 x 100 x 12 = 60,000 cubic feet

25. Application Example Room Design
60,000 cubic feet must be filled 25 times in
an hour. Total air volume needed is
60,000 x 25 = 1,500,000 cubic feet/hour
1,500,000 CFH / 60 min/hr = 25,000 CFM
Validate
60,000/25,000 = 2.4 minutes to fill
60 min/hr divided by 2.4 minutes to fill
equals 25 ACH

26. Worksheet Problem Solving
Five application problems
Please take 15 minutes to work on these.
We will review the answers and how they
were calculated when you are finished.

27. Answer to Question 1 A system contains 20 filters. The filter size is
24x24 and the total system capacity is 40,000 CFM.
What is the velocity in FPM?
Answer – 500 FPM
24x24 = 4 sq. ft. 4 sq. ft. x 20 filters = 80 sq. ft.
40,000 CFM divided by 80 sq. ft. = 500 FPM or
40,000 CFM divided by 20 filters = 2,000 CFM/filter
2,000 CFM divided by 4 sq. ft. = 500 FPM

28. Answer to Question 2 A system has a total capacity of 20,000 CFM and
the velocity through the filter bank is 250 FPM.
How many 24x24 filters does the system contain?
Answer – 20 filters
20,000 CFM divided by 250 FPM = 80 sq. ft.
80 sq. ft. total area divided by 4 sq. ft. = 20 filters or
250 FPM x 4 sq. ft. in a 24x24 = 1,000 CFM/filter
20,000 total CFM divided by 1,000 CFM = 20 filters

29. Answer to Question 3 A specification calls for a maximum of 25 FPM
media velocity through the subject filter at
2,000 CFM. How many square feet of media
must the filter contain?
Answer – 80 sq. ft.
2,000 CFM divided by 25 FPM = 80 sq. ft. media

30. Answer to Question 4 A design engineer wants 10 air changes per hour
In a conditioned space that measures 40x50x12.
How much CFM will be needed?
Answer – 4,000 CFM
40 ft. x 50 ft. x 12 ft = 24,000 cubic feet
24,000 cubic feet changed 10 times = 240,000 CFM
240,000 is the total CFM needed in one hour
240,000 / 60 minutes in an hour = 4,000 CFM

31. Answer to Question 5
An engineer is designing a room with the following parameters
100 ft. x 200 ft. x 12 ft ACH desired are 25
Filters to be 24x24x12 - MERV 14 2,000 CFM each
Maximum media velocity desired is 20 FPM
How much total CFM is needed? Answer – 100,000 CFM
How many filters are needed? Answer – 50 filters
How many sq. ft. of media per filter? Answer – 100 sq. ft.

32. CFM Calculation 100 x 200 x 12 = 240,000 cubic feet of space
240,000 x 25 ACH = 6,000,000 total CFH needed
To convert CFH to CFM you divide CFH by 60min/hr
6,000,000 divided by 60 min/hr = 100,000 CFM

33. Number of Filters Needed
Total CFM needed is 100,000
Each filter is to handle 2,000 CFM
Simply divide 100,000 by 2,000 = 50 filters needed

34. Media Velocity Calculation
Each filter is to handle 2,000 CFM
Maximum media velocity is to be 20 FPM
Simply divide 2,000 CFM by targeted media velocity
of 20 FPM to get minimum media amount
2,000 CFM / 20 FPM = 100 sq. ft. media required

35. End of Presentation
Questions