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YAKINDRA P. TIMILSENA ID No. 111332 AERATION OF STORAGE (MINIMIZING DETERIORATION IN STORAGE)
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Page 1: Aeration of storage structures

YAKINDRA P. TIMILSENA

ID No. 111332

AERATION OF STORAGE(MINIMIZING DETERIORATION IN STORAGE)

Page 2: Aeration of storage structures

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.

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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

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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

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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.

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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.

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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.

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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.

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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

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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

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Step 1 : Select design moisture

The design moisture approximates the equilibrium relative humidity of local climate.

Selected Design moisture for rough rice = 15%

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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

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For Rough rice 15% MC

Dry matter = 100 – 15 = 85 % = 0.85

Dry matter 1 ton release CO2 33.11 g

Dry matter 0.85 ton 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)

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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

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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.

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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.

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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 = 1.987T = Temperature, °CRH = Relative humidity, decimalEXP = “e” to the power, “e” = 2.71828

A,B,C,E,F = equilibrium constants

)]M*BEXP(*C)(T*R

A-EXP[ RH D

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MD =0.15/0.85 = 0.1765 From table : A = 1181.57 , B = 21.733 , C =

35.703T = 30 °C

RH = 0.823 or 82.3%

0.1765)]*21.733EXP(*35.703)(30*1.987

1181.57EXP[RH

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Table2 Chung-Pfost equilibrium constants for grain.

GrainConstant

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

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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%

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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

dayper hoursoperation

air of volumespecific*day) per tonneair of (kg(Q) neededAir

Density of air = 1.177 kg/m 3

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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

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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

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Step 8 : Select fan

(A) area surface

(Q) deliveredair (V)ty air 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

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P = 53.7*0.35441.32 = 13.655/m depth

Pressure drop = 13.655 *4 = 54.62 Pa P = gh (air = 1.177 kg/m3)

Height of bin = 4 m

air of m 73.41.177*9.81

54.62h

g

Ph

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

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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

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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*) W 1000*(0.786power Horse HP

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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.

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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

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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.

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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

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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)L 2π45.5

π

Page 33: Aeration of storage structures

Step 10 : Design the power and controls

Humidistatic controls such as hygrometer and pshychrometer require frequent calibration for accuracy.

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Fans should be selected on the basis of air flow required and static pressure.

Types of Fan

1. Axial-flow fansFan 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

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2. Centrifugal Fans

The fan wheel consists of a hub center, a back plate, the blades, and retaining ring.

Figure 4 Centrifugal Fans

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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. 

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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. 

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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

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……THANK YOU…THANK YOU…