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Mathematics of Air Filtration

Mathematics of Air Filtration. Presented by Rick Peckham 2011 Technical Seminar Dallas, TX – April 13, 2011. 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.)

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Mathematics of Air Filtration

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  1. Mathematics of Air Filtration Presented by Rick Peckham 2011 Technical Seminar Dallas, TX – April 13, 2011

  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 = 2000 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 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 = 2.777 round to 2.78 20x25 = 500 / 144 = 3.472 round to3.47

  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 24 - 20x20 filters. Calculations Convert sq. in. to sq. ft. 20x20 = 400 / 144 which equals 2.777 round to 2.78 sq. ft. 24 x 2.78 = 66.72 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 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. = 113.6 FPM 113.6 is the media velocity for this pleat Bag or pocket filter – 2,000 CFM / 90 sq. ft. equals 22.22 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. = 10.31 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 .50” @ 2000 CFM

  16. Application Example Calculate media velocity 2,000 CFM / 90 sq. ft. = 22.22 FPM The 20x25x2 filter will contain 10 sq. ft. of media area. 10 sq. ft. x 22.22 fpm = 222.2 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 ExampleExisting 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 ExampleExisting 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 ExampleExisting 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 ExampleAlternate 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 HourRoom 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 ExampleRoom 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 ExampleRoom 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

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