This document provides information on aeration of stored grain to minimize deterioration. It defines aeration as forcing air through stored grain to remove heat and moisture. The objectives of grain aeration are outlined, and the typical components of an aeration system are described. Guidelines for the design of aeration systems are then presented, including considerations for airflow rate, fan selection, and air distribution. Steps for designing the aeration of a bulk storage facility are listed and an example design calculation for a rough rice storage is shown.

1. YAKINDRA P. TIMILSENA
ID No. 111332
AERATION OF STORAGE
(MINIMIZING DETERIORATION IN STORAGE)

2. Definition of Aeration
Aeration is a process of forcing air through
stored grain at low flow rates to remove its
heat and moisture and maintain its quality.
It is a very useful storage management tool
which can preserve grain from deterioration,
especially where the moisture content of the
grain is above its safe level.

3. OBJECTIVES OF GRAIN AERATION
• To remove generated heat and water from grain
• To maintain a uniform conditions (temperature,
moisture etc. in the grain bulk)
or equalizing temperature throughout the grain bulk
• Removing or reducing odors from grain
• Removing dryer heat
• Reduce moisture accumulation
• Fumigant application

4. THE AERATION SYSTEM
The components of aeration system
basically consist of the following :
 Aeration ducts
 Air supply duct
 Fan (blower)
 Fan operation control equipment
or controller
 Storage bin

5. Design of Aeration Systems
There are three principal considerations in
the design of aeration systems.
1) airflow rate
2) fan selection
3) air distribution
Automatic controls which are now widely
used may be a part of the system.

6. Airflow rate
 This is the volume of air required to maintain uniform
conditions in the stored bulk by removing the
generated heat and water.
 The recommended rate depends on the purpose of
aeration, the type of grain being aerated, the size
and type of storage structure, and climatic
conditions.

7. Fan selection
 The selection of fan is normally based on the airflow rate
used for a particular grain, the kind of grain handled and
the grain depth.
 These factors determines the resistance of grain to
airflow and the static pressures against which the fan
must deliver the required airflow.
 Two types of fan are used for aeration: centrifugal and the
axial flow fan. Generally, the axial flow fan will deliver
more air than centrifugal fans at a static pressure up to
about 4 inches of water (1,000 Pa). For higher static
pressures, the centrifugal fans are recommended.

8. Air distribution
 This includes the ducts, false floors, etc. which are
used to move the air to the desired points.
 The proper sizing of the ducts, the sizing and
spacing of the openings in the ducts to let the air
move between the duct and aerated grain, the
layout of the duct system are important to maintain
the entering (or exiting) air at an acceptable velocity
and provide uniform airflow through the grain.

9. Design of Aeration of Bulk Storage
There are 10 Steps in the design of aeration of bulk storage
Step 1 : Select design moisture
Step 2 : Calculate the generated heat
Step 3 : Select a design day
Step 4 : Calculate equilibrium humidity
Step 5 : Determine hours of operation per day
Step 6 : Calculate kilogram of air needed per day
Step 7 : Determine air volume and pressure
Step 8 : Select fan
Step 9 : Design the air distribution system
Step 10 : Design the power and controls

10. EXAMPLE FOR DESIGN OF AERATION OF
BULK STORAGE
Given :
Rough rice moisture 15 % (wet basis)
Bin dimension: 6 m x 5m x 4 m (height).
Ambient temperature 30 °C

11. Step 1 : Select design moisture
 The design moisture approximates the
equilibrium relative humidity of local climate.
 Selected Design moisture for rough rice = 15%

12. Step 2 : Calculate the generated heat
 Generated heat should be estimated as a
function of design moisture
 Generated heat can be computed from :
Log(CO2) = AMW –B
For rough rice (MW=15%) : A=0.44 , B=6.08
Log(CO2) = (0.44*15) –6.08 = 0.52
CO2 = 3.311mg/100 gm dry matter
= (0.003311*10000000/1000) gm/ton dry matter
= 33.11 gm /ton dry matter

13.  For Rough rice 15% MC
Dry matter = 100 – 15 = 85 % = 0.85
Dry matter 1 ton release CO2 33.11 g
Dry matter 0.85ton release CO2 28.1435 g
CO2 264 g is equivalent to Heat 2800 kJ
CO2 28.1435 g is equivalent to Heat 298.4917 kJ
Thus, Generated heat = 298.4917 kJ/ton day
Deterioration equation :
C6H12O6 + 6O2 6H2O + 6CO2+ Heat (2800 kJ/mol C6H12O6 )
(180) (192) (108) (264)

14. Table1 Rate of deterioration constants for some common cereal
grains. (To compute CO2 generation)
Grain
A B
10-13.2% 13.3-17% 10-13.2% 13.3-17%
Corn, yellow dent 0.17 0.27 2.00 3.33
Sorghum 0.125 0.32 1.65 4.19
Rough rice 0.21 0.44 3.04 6.08
10-14% 14-17% 10-14% 14-17%
Wheat , soft 0.090 0.36 1.35 5.14

15. Step 3 : Select a design day
 In selecting the design day, local weather data
must be used and as much as possible, these
data should have information on local weather
at least for the last 10 years.
 The wettest month appearing in the data should
be selected.

16. From Figure ,the design day for Jakarta would be in February, it
being the wettest month as shown in the graph.
Fig.5 Relative humidity and temperature data for Jakarta, Indonesia,
latitude 6° 11’ S. The curves represent monthly averages.

17. Step 4 : Calculate equilibrium relative humidity
 Equilibrium relative humidity or reciprocally, grain moisture in
equilibrium with air, may be computed using the information from
Table with the following equations :
MD = E – F * ln [-R*(T+C) ln(RH)]
Where MD = decimal moisture, dry basis
R = universal gas constant = 1.987
T = Temperature, °C
RH = Relative humidity, decimal
EXP = “e” to the power, “e” = 2.71828
A,B,C,E,F = equilibrium constants
)]M*BEXP(*
C)(T*R
A-
EXP[RH D−
+
=

18.  MD =0.15/0.85 = 0.1765
 From table : A = 1181.57 , B = 21.733 , C = 35.703
T = 30 °C
RH = 0.823 or 82.3%
0.1765)]*21.733EXP(*
35.703)(30*1.987
1181.57
EXP[RH −
+
−
=

19. Table2 Chung-Pfost equilibrium constants for grain.
Grain
Constant
A B C E F
Beans, Edible 1334.93 14.964 120.098 .480920 .066826
Corn, Yellow dent 620.56 16.958 30.205 .379212 .058970
Peanut, Kernel 506.65 29.243 33.892 .212966 .034196
Peanut, Pod 1037.19 37.093 12.354 .183212 .026383
Rice, Rough 1181.57 21.733 35.703 .325535 .046015
Sorghum 2185.07 19.644 102.849 .391444 .050970
Soybean 275.11 14.967 24.576 .375314 .066816
Wheat, Durum 1831.40 18.077 112.350 .415593 .055318
Wheat, Hard 1052.01 17.609 50.998 .395155 .056788
Wheat, Soft 1442.54 23.607 35.662 .308163 .042360

20. Step 5 : Determine hours of operation per day
 Hours of operation must be those hours in the design day that fall
below the equilibrium relative humidity.
Hours of operation = 18.75 – 9.5 = 9.25 hours per day
RH =
82.3%

21. Step 6 : Calculate kilogram of air needed per day
 Air needed may be estimated by allowing a 3 °C temperature rise in the
aeration air.
 For a 3 °C rise, the air needed per ton day is calculated as:
kg of air needed = Generated heat / Temperature rise
= 298.4917 / 3
= 99.5 kg of air/ton day
= 99.5 kg of air per ton day * 0.85 m3
/kg of air
9.25 * 60 min
= 0.1524 m3
/ton min
dayperhoursoperation
airofvolumespecific*day)per tonneairof(kg
(Q)neededAir =
Density of air
= 1.177 kg/m3

22. Step 7 : Determine air volume and pressure
 Volume of bin = l*b * h
= 6* 5*4 =120 m3
 From table ; rough rice 1 m3
is occupied by 1.72 ton of grain
 Amount of rough rice = 120 m3
= 69.77 tonnes
1.72 m3
/tonne
Air deliver (Q) = amount of rough rice (tonne) * air needed (m3
/tonne min)
= 69.77 tonne * 0.1524 m3
/tonne min
= 10.633 m3
/min
From step 6

23. Table3 Cubic meters occupied by a tonne of grain
Grain Cubic meters/Tonne
Rough rice 1.72
Maize 1.39
Wheat 1.30
Oats 2.43
Peanuts (Virginia) 4.27
Sorghum 1.37
Barley 1.55

24. Step 8 : Select fan
(A)areasurface
(Q)deliveredair
(V)tyair velociApparent =
= 10.633m3
/min = 0.3544 m/min
6*5
P = 53.7 V1.32
Where P = Pascals of pressure drop in a meter
V = Apparent velocity, in m/min
Static pressure of rough rice :
From step 7

25.  P = 53.7*0.35441.32
= 13.655/m depth
 Pressure drop = 13.655 *4 = 54.62 Pa
 P = ρ gh (ρair = 1.177 kg/m3
)
Height of bin = 4 m
airofm73.4
1.177*9.81
54.62
h ==
g
P
h
ρ
=
Air power = 0.01153 kW * air deliver (m3
/min) * head of air
60 sec/min
= 0.01153 * 10.633 * 4.73
60
= 0.00966 kW

26. Step 8 : Select fan
 Fans should be selected on the basis of air flow required
and static pressure.
 System consists of 2 fans :
Air deliver (Q) = 59.30 / 2 = 29.65 m3
/min
Static pressure = 132.30 Pa ; Pressure loss = 529.2 Pa
Total pressure change = 132.30 + 529.20 = 661.50 Pa
Power = Q∆P = (29.65/60) * 529.2 = 261.51 W = 0.262 kW
From step 7
Step 8 : Select fan

27.  Usually, actual power requirement a fan motor is 3 to 3.5
times for gasoline motors.
 Gasoline motor = 3 * 0.262 = 0.786 kW
1.05
W746
h.p.1
*)W1000*(0.786powerHorse == HP

28. Fan A :
 20 inch diameter,
 3.0 HP,
 2050 RPM,
 Air deliver (Q) = 45.5 m3
/min
 US$ 250
Specifications of possible fan
Fan B :
• 15 inch diameter,
• 4.5 HP,
• 3000 RPM,
• Air deliver (Q) = 55.2 m3
/min
• US$ 400
Fan A is preferable because of the larger wheel, slower
speed, lower power and lower cost.

29. Step 9 : Design the air distribution system
 Duct design consists of two basic velocity constraints.
1) The velocity of the air in the main distribution ducts is :
- For depths of grain ≤ 5 meters
V = 300 to 600 m/min
- For depths of grain > 5 meters
V = 400 to 900 m/min
2) The other velocity constraint refers to the surface area
of the distribution duct.
V ≤ 12 m/min

30.  Q = Av
Where Q = m3
/min of air delivery
A = m2
of area through which air is delivered
V = velocity of delivery, m/min
 Ducts should be a solid distance from the wall equal to
the reciprocal of the depth of grain and may stop at an
equal distance from the wall.
 Ducts are strong when formed in a semicircle.

31. 1) The velocity of the air in the main distribution ducts is :
Q = 45.5 m3
/min (From specification of selected fan)
Height of bin = 4 m select v = 500 m/min
Q = Av
45.5 = 500 * ( D2
/8)
Diameter of duct ; D = 0.48 m or 19.3 inch semi-circle duct
π

32. 2) The velocity of surface area of distribution duct
The length must be long enough to take in the air at the
surface without exceeding 12 m/min velocity.
Q = Av
Q = (2 rL)*v
Length of duct; L = 2.52 m
12*(0.48/2)L2π45.5 =
π

33. Step 10 : Design the power and controls
 Humidistatic controls such as hygrometer and
pshychrometer require frequent calibration for
accuracy.

34. Fans should be selected on the basis of air flow
required and static pressure.
Types of Fan
1. Axial-flow fans
Fan wheels for axial-flow fans are mounted directly on the motor
shaft; the motors are mounted within a tube that serves as the fan
and motor housing
Figure 3 Axial –Flow Fan

35. 2. Centrifugal Fans
The fan wheel consists of a hub center, a back plate, the
blades, and retaining ring.
Figure 4 Centrifugal Fans

36. Selection of a fan in accordance with catalogs that
give the fan characteristics.
GSI Fans
GSI manufactures a complete line of aeration and grain conditioning systems
to help maintain grain quality.
Vane-Axial Fans
Specifications
For applications requiring high airflow at
low static pressures,
GSI offers vane-axial fans in 12" dia.
(3/4 HP) through 42" dia. (40 HP).
The 12" through 28" (15 HP) units
feature a 3450 RPM motor.
Figure 5 Vane –Axial Fans from GSI
Manufacturer.

37. Inline Centrifugal Fans
Specifications
Suited for high static pressures beyond the performance range of
vane-axial fans, the inline centrifugal fan offers the same characteristics of a
regular centrifugal fan, however it is much more economical. Sizes range from
1-1/2 HP to 15 HP, 18" through 28" diameters
Figure 6 Inline Centrifugal Fans from GSI Manufacturer.

38. Designing the Air Distribution System
Type of Duct
Horizontal ducts can be placed at the bottom of the grain bin and may
be used with or without the lateral or branched ducts.
Figure 7. Horizontal ducts for aeration

39. ……THANK YOU…THANK YOU…