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Bernoullis Theorem for Fans

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... are wet-bulb temp and enthalpy. Specific volume are the 'other' slanted ... List the enthalpy, humidity ratio, specific volume and dew point temperature for ... – PowerPoint PPT presentation

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Title: Bernoullis Theorem for Fans


1
Bernoullis 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
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

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
  • Ofpercent floor opening expressed as decimal
  • epvoidage fraction of material expressed as
    decimal (use 0.4 for grains if no better info)

7
ASAE 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 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

9
Shedds Curve (english)
10
Shedds 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 Q4 m3/s

12
FF(pipe)F(exp)F(floor)F(grain)
  • F(pipe)

13
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14
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15
  • f

16
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17
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18
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19
Fexp
20
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21
Ffloor Equ. 2.39
22
  • V Vbin

23
  • Of0.1

24
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25
Fgrain
26
  • 1599 Pa _________ m?

27

28
Using Shedds Curves
  • V0.2 m/s
  • Wheat

29
  • Ftotal 3.2 21.2 2.3 130
  • 157 m

30
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?

31
Moisture and Psychrometrics
  • Core Ag Eng Principles Session IIB

32
Moisture in biological products can be expressed
on a wet basis or dry basis
  • wet basis
  • dry basis (page 273)

33
Standard bushels
  • ASAE Standards
  • Corn weighs 56 lb/bu at 15 moisture wet-basis
  • Soybeans weigh 60 lb/bu at 13.5 moisture
    wet-basis

34
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?

35
  • Remember grain is made up of dry matter H2O
  • The amount of H2O changes, but the amount of dry
    matter in bu is constant.

36
  • Standard bu

37
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38
  • So water removed
  • H2O _at_ 25 - H2O _at_ 13.5

39
Your turn
  • How much water needs to be removed to dry shelled
    corn from 23 (wb) to 15 (wb) if we have 1000 bu?

40
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

41
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

42
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

43
  • Enthalpy 26 BTU/lbda
  • Humidity ratio0.0088 lbH2O/lbda
  • Specific volume 13.55 ft3/lbda
  • Dew point temp 54 F

44
Psychrometric Processes
  • Sensible heating horizontally to the right
  • Sensible cooling horizontally to the left
  • Note that RH changes without changing the
    humidity ratio

45
Psychrometric Processes
  • Evaporative cooling grain drying (p 266)

46
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?

47
Solution
  • _at_ 24C, 68 RH Enthalpy 56 kJ/kgda
  • _at_ 46C Enthalpy 78 kJ/kgda
  • V 0.922 m3/kgda

48
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49
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

50
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51
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

52
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 four T/RH points from EMCs

53
Establishing Exhaust Air RH
  • Draw these points on your psych chart
  • Sketch in a RH curve
  • Where this RH curve intersects your drying
    process line represents the state of the exhaust
    air

54
Sample EMC
55
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?
56
Drying Calculations
57
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.

58
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

59
Step 1
  • How much water must be removed?
  • 2000 bu
  • 20 to 13
  • Now what?

60
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

61
Step 1
62
Step 2
  • How much water can each pound of dry air remove?
  • How do we approach this step?

63
Step 2
  • Find exit conditions from EMC.
  • Plot on psych chart.

0C 32F 64 10C 50F 67 30C 86F 72
64
Step 2
_at_ 52F 68 RH
65
Change in humidity ratio
66
  • Each pound of dry air can remove

67
  • We need to remove 10500 lbH2O.
  • Each lbda removes 0.0023 lbH2O.

68
Step 3
  • Determine the cubic feet of air we need to remove
    necessary water

69
Step 3 Calculations
70
Step 4
  • Determine the fan operating speed
  • How do we approach this?

71
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
72
Step 4
Fgrain
½
PS
Q
6300 cfm
73
  • From cfm of fan and cubic feet of air, determine
    the time needed to dry the soybeans.

74
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75
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.

76
Example 2
  • With heated air, is conserved (not Q)

77
Example 2
  • 2. Determine the relative humidity of the air
    leaving the drier.

78
Example 2
32 40.5 118
78 RH
79
Example 2
  • 3. Determine the amount of propane fuel required
    per hour.

80
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81
Example 2
82
Example 2
  • 4. Determine the amount of fruit residue dried
    per hour.

83
Example 2
  • _at_ 85, 0.15 of every kg is dry matter

84
Example 2
  • Remove 0.85 0.0423

85
Example 2
86
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 -

87
  • 1. What is the necessary fan delivery rate (cfm)?

88
  • 2. What is the approximate total pressure drop
    (in inches of water) required to obtain the
    needed air flow?

89
  • 3. The estimated fan HP based on fan efficiency
    of 65

90
  • 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.
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