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Aeration (Cont’d)

Aeration (Cont’d). Lecture 19. Air Pressure Basics. Measured in lbs/sq inch or psi. A column of air 1 sq ft at sea level weighs 2116 lbs or a ton or 14.7 lbs/sq inch. Atmospheric pressure = 29.92 inches of mercury. The wt. of this column = 14.7 lbs.

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Aeration (Cont’d)

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  1. Aeration (Cont’d) Lecture 19

  2. Air Pressure Basics • Measured in lbs/sq inch or psi. • A column of air 1 sq ft at sea level weighs 2116 lbs or a ton or 14.7 lbs/sq inch. • Atmospheric pressure = 29.92 inches of mercury. The wt. of this column = 14.7 lbs. • So 1 lb/sq inch = 2.036 inches of mercury or 27.74 inches of water.

  3. Static Pressure vs Velocity Pressure • Static pressure = pressure that is non-moving. Air restricted. (Examples: pressure in balloon, tires, air acting against duct internal surface). • Velocity pressure = pressure exerted by moving air. Air not restricted.

  4. 1 psi = 27.74 inches of water 27.74 inches of water = 1 psi 6.8 inches of water = ? (6.8/27.74) x 1 = 0.25 psi.

  5. Static Pressure • Measured using a ‘U” gauge or manometer. • Difference in height of column measured in inches. • Value is positive for push systems of aeration. • Value is negative for pull systems of aeration.

  6. Velocity Pressure • Also measured in inches of water column. • Velocity pressure is influenced by wind velocity or forced air. Example: at 1001 feet per minute velocity (11-12 miles/hr.) = 1/16 inches of water. At 4004 feet per minute (fpm) = 1 inch of water. (16x increase) Velocity pressure is square of the wind velocity.

  7. Static Pressure (SP) vs Velocity Pressure (VP) • Static pressure is resistance to air flow. • Velocity pressure is a function of air flow. • Total pressure (TP) = SP + VP • SP = TP – VP • VP = TP - SP

  8. Calculating Air Velocity and Quantity • Calculate SP and TP using a Pitot tube.

  9. Quantity = cfm/square feet

  10. Relationship between cfm/ft2 vs cfm/bu • cfm/ft2 = grain depth, ft. x (cfm per bu/1.25)

  11. Types of Fans • Axial – air passes through fan in a direction parallel to the fan axis. • Centrifugal – blower wheel located in the center of a spiral housing.

  12. Aeration Fan Selection • Pressure consideration • Airflow consideration • Horse power consideration

  13. Pressure Consideration • Air forced through grain encounters resistance resulting in a pressure drop-a result of friction of the grain mass. • Magnitude of resistance depends on: • Airflow rate • Grain compaction (porosity) – grain type • Distribution of fines • Grain depth Example: 1 cfm/bu moved through 15 ft of shelled corn in bin Air velocity required = cfm/ft2 =15 ft. x 1 cfm/bu/1.25 = 12.0

  14. At 12 cfm/ft2 = static pressure = 0.09 inches.

  15. Total static pressure required = 0.09 inches of water x 15 ft x 1.5 = 2.03 inches of water.

  16. Airflow Considerations • Airflow is affected more by grain depth than grain volume. • Performance data provided by manufacturer. • Example: 36 ft diameter bin of corn. Corn height = 15 ft. Total capacity = 12,215 bushels from Table.

  17. Assuming 1 cfm/bu for 12,215 bushels is equal to an airflow rate of 12,215 cfm. • Assuming a static pressure of 2 inches of water, select the fan from fan performance data (Table).

  18. Fan ID 3, IDs 10-13

  19. Horsepower Considerations • Fan size is indicated in terms of HP. • Actual HP = cfm x static pressure/63.46 x efficiency in %. • 50% efficiency is assumed. • Generally fan HP should be greater than computed HP. • Less SP less the HP required. • Increasing HP has very little effect on cfm/bu.

  20. Air Distribution Considerations • Deliver same airflow rate across all parts of the grain. • Best, perforated floor. • Use of grain spreader.

  21. Above floor ducts: metal or plastic Metal: 6-36 inches in diameter. Durable. Plastic: 8-24 inches in diameter. Inexpensive. Perforated surface area = 5-10% of total duct surface area. Unloading auger location critical.

  22. Aeration Duct Design • Cross section area • Surface area Cross section area = total air volume (cfm)/allowable velocity within the duct (fpm) Minimum total surface area = cfm/allowable exit velocity (fpm) Minimum duct length = minimum total surface area/duct surface area per foot of length

  23. Example Corn in 24 ft diameter bin. Corn ht. 16 ft. 5791 bushels. Minimum cfm/bu = 0.25. Total airflow = 1448 cfm. From table we have selected 2000 fpm as air velocity within the duct. Minimum cross sectional area (ft2) = 1448 cfm/2000 fpm = 0.724 ft2

  24. Minimum total surface area (ft2) = 1448 cfm/30 fpm (exit air velocity) = 48.26 ft2

  25. Minimum duct length (ft) = 48.26/2.51 ft2/ft = 19.23 ft. The bin was of 24 diameter, so this length meets the minimum design requirements.

  26. Cooling Zone

  27. Cooling Time • Aeration cycle = number of hours it takes for the cooling zone to exit the grain surface • Thermocouples = Monitor temperatures throughout bulk to determine cooling zone movement.

  28. Cost of Aeration Cost = 0.7457 x fan HP x hours of operation x electrical cost ($/KwH) (1 HP = 1 KwH)

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