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Storage design for corn yp

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This is my presentation at Asian Institute of Technology Thailand, Graduate Program in Food Engineering

This is my presentation at Asian Institute of Technology Thailand, Graduate Program in Food Engineering

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  • 1. YAKINDRA P. TIMILSENA ID No. 111332 STORAGE DESIGN FOR CORN
  • 2. 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
  • 3. Step 1 : Select design moisture
    • The design moisture approximates the equilibrium relative humidity of local climate.
    • Selected Design moisture for corn = 13%
  • 4. Step 2 : Calculate the generated heat
    • Generated heat should be estimated as a function of design moisture
    • Generated heat can be computed from :
    • Log(CO 2 ) = AM W –B
    • For corn (M W =13%) : A=0.17 , B=2.00
    • Log(CO 2 ) = (0.17*13) –2.00 = 0. 21
    • CO 2 = 1.6218 mg/100 gm dry matter
    • = (0.0016218*10000000/1000) gm/ton dry matter
    • = 16.22 gm /ton dry matter
  • 5.
    • For Corn with 13% MC
    • Dry matter = 100 – 13 = 87 % = 0.87
    • Dry matter 1 ton release CO 2 16.22 g
    • Dry matter 0.87 ton release CO 2 14.1114 g
    • CO 2 264 g is equivalent to Heat 2800 kJ
    • CO 2 14.1114 g is equivalent to Heat 149.6664 kJ
    • Thus, Generated heat = 149.6664 kJ/ton day
    Deterioration equation : C 6 H 12 O 6 + 6O 2 6H 2 O + 6CO 2 + Heat (2800 kJ/mol C 6 H 12 O 6 ) (180) (192) (108) (264)
  • 6. Table1 Rate of deterioration constants for some common cereal grains. ( To compute CO 2 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
  • 7. 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.
  • 8.
    • 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.
  • 9. 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 :
    • M D = E – F * ln [-R*(T+C) ln(RH)]
    • Where M D = 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
  • 10.
    • M D =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)] * 16.958 EXP( * 30.205) (30 * 1.987 620.56 EXP[ RH    
  • 11. Table2 Chung-Pfost equilibrium constants for grain. Grain Constant 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
  • 12. 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%
  • 13. 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 m 3 /kg of air
    • 6.5 * 60 min
    • = 0.1087m 3 /ton min
    Density of air = 1.177 kg/m 3
  • 14. Step 7 : Determine air volume and pressure
    • Volume of bin = l*b * h
    • = 6 * 5*4 =120 m 3
    • From table ; Maize/corn 1 m 3 is occupied by 1.39 ton of grain
    • Amount of corn = 120 m 3 = 86.33 tonnes
    • 1.39 m 3 /tonne
    • Air deliver (Q) = amount of corn (tonne) * air needed (m 3 /tonne min)
    • = 86.33 tonne * 0.1087 m 3 /tonne min
    • = 9.384 m 3 /min
    From step 6
  • 15. 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
  • 16. Step 8 : Select fan = 59.30 m 3 /min = 1.98 m/min 6*5 P = 53.7 V 1.32 Where P = Pascals of pressure drop in a meter V = Apparent velocity, in m/min Static pressure of rough rice : From step 7
  • 17.
    • P = 53.7*1.98 1.32 = 132.30 /m depth
    • Pressure drop = 132.30 *4 = 529.2Pa
    • P =  gh (  air = 1.177 kg/m 3 )
    Height of bin = 4 m Air power = 0.01153 kW * air deliver (m 3 /min) * head of air 60 sec/min = 0.01153 * 59.30 * 45.83 60 = 0.522 kW
  • 18. 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 m 3 /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
  • 19.
    • Usually, actual power requirement a fan motor is 3 to 3.5 times for gasoline motors.
    • Gasoline motor = 3 * 0.262 = 0.786 kW
    HP
  • 20.
    • Fan A :
    • 20 inch diameter,
    • 3.0 HP,
    • 2050 RPM,
    • Air deliver (Q) = 45.5 m 3 /min
    • US$ 250
    Specifications of possible fan
    • Fan B :
    • 15 inch diameter,
    • 4.5 HP,
    • 3000 RPM,
    • Air deliver (Q) = 55.2 m 3 /min
    • US$ 400
    Fan A is preferable because of the larger wheel, slower speed, lower power and lower cost.
  • 21. 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
  • 22.
    • Q = Av Where Q = m 3 /min of air delivery
    • A = m 2 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.
  • 23.
    • 1) The velocity of the air in the main distribution ducts is :
    • Q = 45.5 m 3 /min (From specification of selected fan)
    • Height of bin = 4 m select v = 500 m/min
    • Q = Av
    • 45.5 = 500 * ( D 2 /8)
    • Diameter of duct ; D = 0.48 m or 19.3 inch semi-circle duct
  • 24.
    • 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
  • 25. Step 10 : Design the power and controls
    • Humidistatic controls such as hygrometer and pshychrometer require frequent calibration for accuracy.
  • 26. … THANK YOU…

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