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Bernoulli’s Theorem for Fans PE Review Session VIB – section 1

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Fan and Bin 1 2 3

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static pressure velocity head total pressure

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Power

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F total =F pipe +F expansion +F floor +F grain F pipe =f (L/D) (V 2 /2g) for values in pipe F expansion = (V 1 2 – V 2 2 ) / 2g V 1 is velocity in pipe V 2 is velocity in bin V 1 >> V 2 so equation reduces to V 1 2 /2g

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F floor Equation 2.38 p. 29 (4 th edition) for no grain on floor Equation 2.39 p. 30 (4 th edition) for grain on floor O f =percent floor opening expressed as decimal ε p =voidage fraction of material expressed as decimal (use 0.4 for grains if no better info)

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ASABE Standards - graph for F floor

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F grain Equation 2.36 p. 29 (C f = 1.5) A and b from standards or Table 2.5 p. 30 Or use Shedd’s curves (Standards) X axis is pressure drop/depth of grain Y axis is superficial velocity (m 3 /(m 2 s) Multiply pressure drop by 1.5 for correction factor Multiply by specific weight of air to get F in m or f

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Shedd’s Curve (english)

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Shedd’s curves (metric)

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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 m 2. Find the static and total pressure when Q=4 m 3 /s

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F=F(pipe)+F(exp)+F(floor)+ F(grain) F(pipe)=

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ff

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F exp

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F floor Equ. 2.39

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V = V bin =

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O f =0.1

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F grain

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1599 Pa = _________ m?

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Using Shedd’s Curves V=0.2 m/s Wheat

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F total = = 157 m

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Problem 2.4 (page 45) Air (21C) at the rate of 0.1 m 3 /(m 2 s) is to be moved vertically through a crib of shelled corn 1.6 m deep. The area of the floor is 12 m 2 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%?

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Moisture and Psychrometrics Core Ag Eng Principles Session IIB

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

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Standard bushels ASABE Standards Corn weighs 56 lb/bu at 15% moisture wet-basis Soybeans weigh 60 lb/bu at 13.5% moisture wet-basis

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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%?

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Remember grain is made up of dry matter + H 2 O The amount of H 2 O changes, but the amount of dry matter in bu is constant.

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Standard bu

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So water removed = H 2 25% %

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

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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

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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

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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

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Enthalpy = 26 BTU/lb da Humidity ratio= lb H2O /lb da Specific volume = ft 3 /lb da Dew point temp = 54 F

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

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Psychrometric Processes Evaporative cooling = grain drying (p 266)

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Example A grain dryer requires 300 m 3 /min of 46C air. The atmospheric air is at 24C and 68% RH. How much power must be supplied to heat the air?

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24C, 68% RH: Enthalpy = 56 kJ/kg 46C: Enthalpy = 78 kJ/kg da V = m 3 /kg da

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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

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Equilibrium Moisture Curves Establish second point on the evaporative cooling line – i.e. can’t 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 won’t “release” any more water

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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 three T/RH points from EMCs (the fourth one is typically out of the temperature range)

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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

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Sample EMC

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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?

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Drying Calculations

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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 cfm at ½” H 2 O static pressure. The bin is 26’ in diameter and outside air (60 F, 30% RH) is being blown over the soybeans.

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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

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

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Step 1 Std bu = m w = 0.135(60 lb) = 8.1 lb H 2 O m d = m t – m w = 60 – 8.1 = 51.9 lb 13%:

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Step 1

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

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Step 2 Find exit conditions from EMC. Plot on psych chart. 0C = 32F = 64% 10C = 50F = 67% 30C = 86F = 72%

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Step 52F – 68% RH

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Change in humidity ratio

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Each pound of dry air can remove

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We need to remove 10,500 lb H2O. Each lb da removes lb H2O.

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

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Step 3 Calculations

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Step 4 Determine the fan operating speed How do we approach this?

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Step 4 Main term in F is F grain Airflow (cfm/ft 2 ) Pressure drop (“H 2 O/ft) x depth x CF

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Step 4 ½ F grain 6300 cfm Q PSPS

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From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.

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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 m 3 /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.

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

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

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Example % RH

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Example 2 3. Determine the amount of propane fuel required per hour.

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Example 2

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4. Determine the amount of fruit residue dried per hour.

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Example 85%, 0.15 of every kg is dry matter

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Example 2 Remove 0.85 – =

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Example 2

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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 -

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

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

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3. The estimated fan HP based on fan efficiency of 65%

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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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