Core Principles

1
Core Principles

• Bernoullis theorem for Fans
• Friction Loss through Grain
• Moisture Content
• Psychrometrics
• Equilibrium Moisture Curves

2
Bernoullis 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
FtotalFpipeFexpansionFfloorFgrain

• Fpipef (L/D) (V2/2g) for values in pipe
• Fexpansion (V12 V22) / 2g
• V1 is velocity in pipe
• V2 is velocity in bin
• V1 gtgt V2 so equation reduces to
• V12/2g

8
Ffloor

• Equation 2.38 p. 29 (4th edition) for no grain on
  floor
• Equation 2.39 p. 30 (4th edition) for grain on
  floor
• Ofpercent floor opening expressed as decimal
• epvoidage 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 Shedds 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
Shedds Curve (english)
12
Shedds 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 0.4.
  The area of the bin floor is 20 m2. Find the
  static and total pressure when Q4 m3/s

14
Fan and Bin
3

10 opening
5 m
1 m
2
1
0.5m ID
plenum
Q
Q
15
FF(pipe)F(exp)F(floor)F(grain)

• F(pipe)

16
(No Transcript)
17
(No Transcript)
18

• f

19
(No Transcript)
20
(No Transcript)
21
(No Transcript)
22
Fexp
23
(No Transcript)
24
Ffloor Equ. 2.39
25

• V Vbin

26

• Of0.1

27
(No Transcript)
28
Fgrain
29

• 1599 Pa _________ m?

30

31
Using Shedds Curves

• V0.2 m/s
• Wheat

32

• Ftotal 3.2 m 21.2 m 2.3 m 130 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

• wet 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

40
(No Transcript)
41

• So water removed
• H2O _at_ 25 - H2O _at_ 13.5

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 ratio0.0088 lbH2O/lbda
• Specific volume 13.55 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

• _at_ 24C, 68 RH Enthalpy 56 kJ/kgda
• _at_ 46C Enthalpy 78 kJ/kgda
• V 0.922 m3/kgda

51
(No Transcript)
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

53
(No Transcript)
54
Equilibrium Moisture Curves

• Establish second point on the evaporative cooling
  line i.e. cant 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 wont 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

• _at_ 24C, 68 RH Enthalpy 56 kJ/kgda
• _at_ 46C Enthalpy 78 kJ/kgda
• V 0.922 m3/kgda

59
Establishing Exhaust Air RH

• Sketch in a RH curve

60
Solution

• _at_ 24C, 68 RH Enthalpy 56 kJ/kgda
• _at_ 46C Enthalpy 78 kJ/kgda
• V 0.922 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

• _at_ 24C, 68 RH Enthalpy 56 kJ/kgda
• _at_ 46C Enthalpy 78 kJ/kgda
• V 0.922 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 5140-9000 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 _at_ 0.135
• mw 0.135(60 lb) 8.1 lb H2O
• md mt mw 60 8.1 51.9 lbdm
• _at_ 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
_at_ 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 0.0023 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
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
½
Fan Curve
PS
Q
6300 cfm
81

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

82
(No Transcript)
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
32 40.5 118
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

• _at_ 85, 0.15 of every kg is dry matter

91
Example 2

• Remove 0.85 0.0423

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.