Presentation on theme: "Bernoulli’s Theorem for Fans PE Review Session VIB – section 1."— Presentation transcript:
Bernoulli’s Theorem for Fans PE Review Session VIB – section 1
Fan and Bin 1 2 3
static pressure velocity head total pressure
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
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)
ASABE Standards - graph for F floor
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
Shedd’s Curve (english)
Shedd’s curves (metric)
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
F=F(pipe)+F(exp)+F(floor)+ F(grain) F(pipe)=
F floor Equ. 2.39
V = V bin =
O f =0.1
1599 Pa = _________ m?
Using Shedd’s Curves V=0.2 m/s Wheat
F total = 3.2 + 21.2 + 2.3 + 130 = 157 m
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%?
Moisture and Psychrometrics Core Ag Eng Principles Session IIB
Moisture in biological products can be expressed on a wet basis or dry basis wet basis dry basis (page 273)
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
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%?
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.
So water removed = H 2 O @ 25% - H2O @ 13.5%
Your turn: How much water needs to be removed to dry shelled corn from 23% (wb) to 15% (wb) if we have 1000 bu?
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
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
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
Enthalpy = 26 BTU/lb da Humidity ratio=0.0088 lb H2O /lb da Specific volume = 13.55 ft 3 /lb da Dew point temp = 54 F
Psychrometric Processes Sensible heating – horizontally to the right Sensible cooling – horizontally to the left Note that RH changes without changing the humidity ratio
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?
Solution @ 24C, 68% RH: Enthalpy = 56 kJ/kg da @ 46C: Enthalpy = 78 kJ/kg da V = 0.922 m 3 /kg da
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
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
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)
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
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?
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 5140-9000 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.
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
Step 1 How much water must be removed? 2000 bu 20% to 13% Now what?
Step 1 Std bu = 60 lb @ 0.135 m w = 0.135(60 lb) = 8.1 lb H 2 O m d = m t – m w = 60 – 8.1 = 51.9 lb dm @ 13%:
Step 2 How much water can each pound of dry air remove? How do we approach this step?
We need to remove 10,500 lb H2O. Each lb da removes 0.0023 lb H2O.
Step 3 Determine the cubic feet of air we need to remove necessary water
Step 3 Calculations
Step 4 Determine the fan operating speed How do we approach this?
Step 4 Main term in F is F grain Airflow (cfm/ft 2 ) 50 30 15 10 Pressure drop (“H 2 O/ft) 0.5 0.23 0.09 0.05 x depth x CF
Step 4 ½ F grain 6300 cfm Q PSPS
From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.
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.
Example 2 With heated air, is conserved (not Q)
Example 2 2. Determine the relative humidity of the air leaving the drier.
Example 2 32 40.5 118 78% RH
Example 2 3. Determine the amount of propane fuel required per hour.
4. Determine the amount of fruit residue dried per hour.
Example 2 @ 85%, 0.15 of every kg is dry matter
Example 2 Remove 0.85 – 0.0423 =
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 -
1. What is the necessary fan delivery rate (cfm)?
2. What is the approximate total pressure drop (in inches of water) required to obtain the needed air flow?
3. The estimated fan HP based on fan efficiency of 65%
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.