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Bernoulli’s Theorem for Fans

Bernoulli’s Theorem for Fans. PE Review Session VIB – section 1. Fan and Bin. 3. 2. 1. static pressure. velocity head. total pressure. Power. F total =F pipe +F expansion +F floor +F grain. F pipe =f (L/D) (V 2 /2g) for values in pipe F expansion = (V 1 2 – V 2 2 ) / 2g

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Bernoulli’s Theorem for Fans

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  1. Bernoulli’s Theorem for Fans PE Review Session VIB – section 1

  2. Fan and Bin 3 2 1

  3. static pressure velocity head total pressure

  4. Power

  5. Ftotal=Fpipe+Fexpansion+Ffloor+Fgrain • Fpipe=f (L/D) (V2/2g) for values in pipe • Fexpansion= (V12 – V22) / 2g • V1 is velocity in pipe • V2 is velocity in bin • V1 >> V2 so equation reduces to • V12/2g

  6. Ffloor • Equation 2.38 p. 29 (4th edition) for no grain on floor • Equation 2.39 p. 30 (4th edition) for grain on floor • Of=percent floor opening expressed as decimal • εp=voidage fraction of material expressed as decimal (use 0.4 for grains if no better info)

  7. ASABE Standards - graph for Ffloor

  8. Fgrain • Equation 2.36 p. 29 (Cf = 1.5) • A and b from standards or Table 2.5 p. 30 • Or use Shedd’s curves (Standards) • X axis is pressure drop/depth of grain • Y axis is superficial velocity (m3/(m2s) • Multiply pressure drop by 1.5 for correction factor • Multiply by specific weight of air to get F in m or f

  9. Shedd’s Curve (english)

  10. Shedd’s curves (metric)

  11. Example • Air is to be forced through a grain drying bin similar to that shown before. The air flows through 5 m of 0.5 m diameter galvanized iron conduit, exhausts into a plenum below the grain, passes through a perforated metal floor (10% openings) and is finally forced through a 1 m depth of wheat having a void fraction of 0.4. The area of the bin floor is 20 m2. Find the static and total pressure when Q=4 m3/s

  12. F=F(pipe)+F(exp)+F(floor)+F(grain) • F(pipe)=

  13. f

  14. Fexp

  15. Ffloor Equ. 2.39

  16. V = Vbin =

  17. Of=0.1

  18. Fgrain

  19. 1599 Pa = _________ m?

  20. Using Shedd’s Curves • V=0.2 m/s • Wheat

  21. Ftotal = 3.2 + 21.2 + 2.3 + 130 • = 157 m

  22. Problem 2.4 (page 45) • Air (21C) at the rate of 0.1 m3/(m2 s) is to be moved vertically through a crib of shelled corn 1.6 m deep. The area of the floor is 12 m2 with an opening percentage of 10% and the connecting galvanized iron pipe is 0.3 m in diameter and 12 m long. What is the power requirement, assuming the fan efficiency to be 70%?

  23. Moisture and Psychrometrics Core Ag Eng Principles Session IIB

  24. Moisture in biological products can be expressed on a wet basis or dry basis wet basis dry basis (page 273)

  25. Standard bushels • ASABE Standards • Corn weighs 56 lb/bu at 15% moisture wet-basis • Soybeans weigh 60 lb/bu at 13.5% moisture wet-basis

  26. Use this information to determine how much water needs to be removed to dry grain • We have 2000 bu of soybeans at 25% moisture (wb). How much water must be removed to store the beans at 13.5%?

  27. Remember grain is made up of dry matter + H2O • The amount of H2O changes, but the amount of dry matter in bu is constant.

  28. Standard bu

  29. So water removed = H2O @ 25% - H2O @ 13.5%

  30. Your turn: • How much water needs to be removed to dry shelled corn from 23% (wb) to 15% (wb) if we have 1000 bu?

  31. Psychrometrics • If you know two properties of an air/water vapor mixture you know all values because two properties establish a unique point on the psych chart • Vertical lines are dry-bulb temperature

  32. Psychrometrics • Horizontal lines are humidity ratio (right axis) or dew point temp (left axis) • Slanted lines are wet-bulb temp and enthalpy • Specific volume are the “other” slanted lines

  33. Your turn: • List the enthalpy, humidity ratio, specific volume and dew point temperature for a dry bulb temperature of 70F and a wet-bulb temp of 60F

  34. Enthalpy = 26 BTU/lbda • Humidity ratio=0.0088 lbH2O/lbda • Specific volume = 13.55 ft3/lbda • Dew point temp = 54 F

  35. Psychrometric Processes • Sensible heating – horizontally to the right • Sensible cooling – horizontally to the left • Note that RH changes without changing the humidity ratio

  36. Psychrometric Processes • Evaporative cooling = grain drying (p 266)

  37. Example • A grain dryer requires 300 m3/min of 46C air. The atmospheric air is at 24C and 68% RH. How much power must be supplied to heat the air?

  38. Solution @ 24C, 68% RH: Enthalpy = 56 kJ/kgda @ 46C: Enthalpy = 78 kJ/kgda V = 0.922 m3/kgda

  39. Equilibrium Moisture Curves • When a biological product is in a moist environment it will exchange water with the atmosphere in a predictable way – depending on the temperature/RH of the moist air surrounding the biological product. • This information is contained in the EMC for each product

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