1 / 89

Bernoulli’s Theorem for Fans

This review session covers section 1 of the PE exam, focusing on Bernoulli's theorem for fans. Topics include static pressure, velocity head, total pressure, power, and calculations using Shedd's curves.

beaudin
Download Presentation

Bernoulli’s Theorem for Fans

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related