Ftotal=Fpipe+Fexpansion+Ffloor+Fgrain
Fpipe=f (L/D) (V2/2g) for values in pipe
Fexpansion= (V12 – V2
2) / 2g V1 is velocity in pipe
V2 is velocity in bin
V1 >> V2 so equation reduces to
V12/2g
Ffloor
Equation 2.38 p. 29 (4th edition) for no grain on floor
Equation 2.39 p. 30 (4th edition) for grain on floor Of=percent floor opening expressed as decimal
εp=voidage fraction of material expressed as
decimal (use 0.4 for grains if no better info)
Fgrain
Equation 2.36 p. 29 (Cf = 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 (m3/(m2s) Multiply pressure drop by 1.5 for
correction factor Multiply by specific weight of air to get F
in m or f
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 m2. Find the static and total pressure when Q=4 m3/s
Problem 2.4 (page 45) Air (21C) at the rate of 0.1 m3/(m2 s) is
to be moved vertically through a crib of shelled corn 1.6 m deep. The area of the floor is 12 m2 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 in biological products can be expressed on a wet basis or dry basis
wet basis
dry basis (page 273)d
m
dm
m
W
WM
)W(W
Wm
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 grainWe 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 + H2O
The amount of H2O changes, but the amount of dry matter in bu is constant.
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/lbda
Humidity ratio=0.0088 lbH2O/lbda
Specific volume = 13.55 ft3/lbda
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 m3/min of
46C air. The atmospheric air is at 24C and 68% RH. How much power must be supplied to heat the air?
Equilibrium Moisture CurvesWhen 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” H2O 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 1Std bu = 60 lb @ 0.135mw = 0.135(60 lb) = 8.1 lb H2O
md = mt – mw = 60 – 8.1 = 51.9 lbdm
@ 13%:
7.76lbm
6.750.13mm
51.9m
m0.13
w
ww
w
w
Step 2Find exit conditions from EMC.Plot on psych chart.
0C = 32F = 64%10C = 50F = 67%30C = 86F = 72%
We need to remove 10,500 lbH2O.
Each lbda removes 0.0023 lbH2O.
OH
daOHda
2
2 0.0023lb
1lb10500lbb4,565,217l
Step 4Main term in F is Fgrain
Airflow (cfm/ft2)50301510
Pressure drop (“H2O/ft)0.5
0.230.090.05
x depth x CF
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 m3/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 2With heated air, is conserved (not Q)
m
s
m7.65
1.125m
kg
s
kg6.8Q
s
kg6.8
0.875m
kg
s
m5.95m
3
3pt
3
3
2
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 -
2. What is the approximate total pressure drop (in inches of water) required to obtain the needed air flow?