1. Core Principles Bernoulli’s theorem for Fans
Friction Loss through Grain
Moisture Content
Psychrometrics
Equilibrium Moisture Curves

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

3. Fan and Bin 3 2 1

4. static
pressure
velocity
head
total
pressure

5. Power P = Power (total) PT = Total Pressure P = Power (static)
PS = Static Pressure

6. Friction Loss through Grain

7. 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

8. Ffloor 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)
Ffloor

9. ASABE Standards - graph for Ffloor

10. 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

11. Shedd’s Curve (english)

12. Shedd’s curves (metric)

13. 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 The area of the bin floor is 20 m2. Find the static and total pressure when Q=4 m3/s

14. Fan and Bin 3
10% opening
5 m
1 m 2 1 Q
0.5m ID
plenum Q

15. F=F(pipe)+F(exp)+F(floor)+ F(grain)

18. f

22. Fexp

24. Ffloor Equ. 2.39

25. V = Vbin =

26. Of=0.1

28. Fgrain

29. 1599 Pa = _________ m?

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

32. Ftotal = 3.2 m m m m
Ftotal = 157 m

33. Your Turn: 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%?

34. Moisture and Psychrometrics
Core Ag Eng Principles Session IIB

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

36. 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

37. 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%?

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

39. Standard bu

41. So water removed =
25% %

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

43. 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

44. 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

45. 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

46. Enthalpy = 26 BTU/lbda
Humidity ratio= lbH2O/lbda
Specific volume = ft3/lbda
Dew point temp = 54 F

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

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

49. 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?

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

52. 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

54. Equilibrium Moisture Curves
Establish second point on the evaporative cooling line – i.e. can’t remove enough water from the product to saturate the air under all conditions – sometimes the exhaust air is at a lower RH because the product won’t “release” any more water

55. Establishing Exhaust Air RH
Select EMC for product of interest
On Y axis – draw horizontal line at the desired final moisture content (wb) of product
Find the three T/RH points from EMCs (the fourth one is typically out of the temperature range)

56. Sample EMC

57. Establishing Exhaust Air RH
Draw these points on your psych chart

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

59. Establishing Exhaust Air RH
“Sketch” in a RH curve

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

61. Establishing Exhaust Air RH
Where this RH curve intersects your drying process line represents the state of the exhaust air

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

63. We are drying corn to 15% wb; with natural ventilation using outside air at 25C and 70% RH. What will be the Tdb and RH of the exhaust air?

64. Drying Calculations

65. Example problem
How long will it take to dry 2000 bu of soybeans from 20% mc (wb) to 13% mc (wb) with a fan which delivers cfm at ½” H2O static pressure. The bin is 26’ in diameter and outside air (60 F, 30% RH) is being blown over the soybeans.

66. Steps to work drying problem
Determine how much water needs to be removed (from moisture content before and after; total amount of product to be dried)
Determine how much water each pound of dry air can remove (from psychr chart; outside air – is it heated, etc., and EMC)
Calculate how many cubic feet of air is needed
Determine fan operating CFM
From CFM, determine time needed to dry product

67. Step 1
How much water must be removed?
2000 bu
20% to 13%
Now what?

68. Step 1 Std bu = 60 lb @ 0.135 mw = 0.135(60 lb) = 8.1 lb H2O
md = mt – mw = 60 – 8.1 = 51.9 lbdm
@ 13%:

69. Step 1

70. Step 2 How much water can each pound of dry air remove?
How do we approach this step?

71. Step 2 Find exit conditions from EMC. Plot on psych chart.
0C = 32F = 64%
10C = 50F = 67%
30C = 86F = 72%

72. Step 2
@ 52F – 68% RH

73. Change in humidity ratio

74. Each pound of dry air can remove

75. We need to remove 10,500 lbH2O.
Each lbda removes lbH2O.

76. Step 3
Determine the cubic feet of air we need to remove necessary water

77. Step 3 Calculations

78. Step 4
Determine the fan operating speed
How do we approach this?

79. Pressure drop (“H2O/ft)
Step 4
Main term in F is Fgrain
Airflow (cfm/ft2) 50 30 15 10
Pressure drop (“H2O/ft)
0.5
0.23
0.09
0.05
x depth x CF

80. Step 4
Fgrain
System Characterisitc Curve
Fan Curve PS Q
6300 cfm

81. From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.

83. Example 2
Ambient air at 32C and 20% RH is heated to 118 C in a fruit residue dryer. The flow of ambient air into the propane heater is at 5.95 m3/sec. The drying is to be carried out from 85% to 22% wb. The air leaves the drier at 40.5C.
Determine the airflow rate of the heated air.

84. Example 2
With heated air, is conserved (not Q)

85. Example 2
2. Determine the relative humidity of the air leaving the drier.

86. Example 2
78% RH

87. Example 2
3. Determine the amount of propane fuel required per hour.

88. Example 2

89. Example 2
4. Determine the amount of fruit residue dried per hour.

90. Example 2
@ 85%, 0.15 of every kg is dry matter

91. Example 2
Remove 0.85 – =

92. Example 2

93. Your Turn:
A grain bin 26’ in diameter has a perforated floor over a plenum chamber. Shelled field corn will be dried from an initial mc of 24% to 14% (wb). Batch drying (1800 std. bu/batch) will be used with outside air (55F, RH 70%) that has been heated 10F before being passed through the corn. To dry the corn in 1 week -

94. 1. What is the necessary fan delivery rate (cfm)?

95. 2. What is the approximate total pressure drop (in inches of water) required to obtain the needed air flow?

96. 3. The estimated fan HP based on fan efficiency of 65%

97. 4. If the drying air is heated by electrical resistance elements and the power costs is $0.065/KWH, calculate the cost of heating energy per standard bushel.