YAKINDRA P. TIMILSENA

ID No. 111332

STORAGE DESIGN FOR CORN

DESIGN OF AERATION OF BULK STORAGE

Given :Corn moisture 13 % (wet basis)Bin dimension: 6 m x 5m x 4 m (height).Ambient temperature 30 °C

Step 1 : Select design moisture

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

Selected Design moisture for corn = 13%

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 corn (MW=13%) : A=0.17 , B=2.00

Log(CO2) = (0.17*13) –2.00 = 0. 21

CO2 = 1.6218 mg/100 gm dry matter

= (0.0016218*10000000/1000) gm/ton dry matter

= 16.22 gm /ton dry matter

For Corn with 13% MC

Dry matter = 100 – 13 = 87 % = 0.87

Dry matter 1 ton release CO2 16.22 g

Dry matter 0.87 ton release CO2 14.1114 g

CO2 264 g is equivalent to Heat 2800 kJ

CO2 14.1114 g is equivalent to Heat 149.6664 kJ

Thus, Generated heat = 149.6664 kJ/ton day

Deterioration equation :

C6H12O6 + 6O2 6H2O + 6CO2 + Heat (2800 kJ/mol C6H12O6 ) (180) (192) (108) (264)

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

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.

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.

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

MD =0.13/0.87 = 0.1494 From table : A = 620.56 , B = 16.958 , C = 30.205

T = 30 °C

RH = 0.6625 or 66.25%

0.1494)]*16.958EXP(*30.205)(30*1.987

620.56EXP[RH

Table2 Chung-Pfost equilibrium constants for grain.

GrainConstant

A B C E F

Beans, Edible 1334.93 14.964 120.098 .480920 .066826

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

Corn, Yellow dent 620.56 16.958 30.205 .379212 .058970

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

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 = 17.75 – 11.25 = 6.5 hours per day

RH = 66.25%

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 = 149.6664 / 3 = 49.88 kg of air/ton day

= 49.88 kg of air per ton day * 0.85 m3/kg of air 6.5 * 60 min

= 0.1087m3/ton min

dayper hoursoperation

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

Density of air = 1.177 kg/m 3

Step 7 : Determine air volume and pressure

Volume of bin = l*b * h = 6* 5*4 =120 m3

From table ; Maize/corn 1 m3 is occupied by 1.39 ton of grain

Amount of corn = 120 m3 = 86.33 tonnes 1.39 m3/tonne

Air deliver (Q) = amount of corn (tonne) * air needed (m3/tonne min)

= 86.33 tonne * 0.1087 m3/tonne min = 9.384 m3/min

From step 6

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

Step 8 : Select fan

(A) area surface

(Q) deliveredair (V)ty air velociApparent

= 59.30 m3/min = 1.98 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

P = 53.7*1.981.32 = 132.30 /m depth

Pressure drop = 132.30 *4 = 529.2Pa P = gh (air = 1.177 kg/m3)

Height of bin = 4 m

air of m 83.541.177*9.81

529.2h

g

Ph

Air power = 0.01153 kW * air deliver (m3/min) * head of air

60 sec/min = 0.01153 * 59.30 * 45.83

60 = 0.522 kW

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

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

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.

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

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.

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

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

Step 10 : Design the power and controls

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

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