bernoulli’s theorem for fans
DESCRIPTION
Bernoulli’s Theorem for Fans. PE Review Session VIB – section 1. Fan and Bin. 3. 2. 1. static pressure. velocity head. total pressure. Power. 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 - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/1.jpg)
Bernoulli’s Theorem for FansPE Review Session VIB – section 1
![Page 2: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/2.jpg)
Fan and Bin
γ
Ph
γ
Ph
2g
v
γ
PhFW
2g
v
γ
Ph
33
11
233
3
211
1
12
3
![Page 3: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/3.jpg)
Fγ
P
2g
v
γ
PW
FW
0vv
T222
31
staticpressure
velocityhead
total pressure
![Page 4: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/4.jpg)
Power
s
s
T
T
e
QPP
or
e
QPP
![Page 5: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/5.jpg)
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
![Page 6: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/6.jpg)
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)
![Page 7: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/7.jpg)
ASABE Standards - graph for Ffloor
![Page 8: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/8.jpg)
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
![Page 9: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/9.jpg)
Shedd’s Curve (english)
![Page 10: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/10.jpg)
Shedd’s curves (metric)
![Page 11: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/11.jpg)
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
![Page 12: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/12.jpg)
F=F(pipe)+F(exp)+F(floor)+F(grain) F(pipe)=
![Page 13: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/13.jpg)
g2
v
D
LfF
2
pipe
![Page 14: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/14.jpg)
pipe
pipe A
QV
![Page 15: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/15.jpg)
f
Dv
andD
Re
Re
![Page 16: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/16.jpg)
Re
![Page 17: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/17.jpg)
)(
103
5.01000
115.0
4
moodyf
mmmm
mm
D
![Page 18: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/18.jpg)
pipeF
f
![Page 19: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/19.jpg)
Fexp
g
vvF
2
22
21
exp
![Page 20: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/20.jpg)
m
sm
sm
F 2.2181.92
04.20
2
2
exp
![Page 21: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/21.jpg)
Ffloor Equ. 2.39
g
ov
msPa
pf
2
2
2
071.1
![Page 22: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/22.jpg)
V = Vbin = s
m
ms
m
A
Q
bin
2.020
4
2
3
![Page 23: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/23.jpg)
Of=0.1
4.0p
![Page 24: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/24.jpg)
m
sm
mkgmsPa
Ffloor 3.281.9202.1
4.01.02.0
071.1
22
2
2
![Page 25: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/25.jpg)
Fgrain
Pa
sm
msm
P
bV
cVa
L
PF fwheat
1599
2.077.81ln
15.12.0107.2
1ln2
4
2
![Page 26: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/26.jpg)
1599 Pa = _________ m?
![Page 27: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/27.jpg)
m
sm
mkg
mN
gmN
13581.9202.1
15991599
23
22
![Page 28: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/28.jpg)
Using Shedd’s CurvesV=0.2 m/sWheat
L
P
![Page 29: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/29.jpg)
Ftotal = 3.2 + 21.2 + 2.3 + 130
= 157 m
![Page 30: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/30.jpg)
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%?
![Page 31: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/31.jpg)
Moisture and PsychrometricsCore Ag Eng Principles Session IIB
![Page 32: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/32.jpg)
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
![Page 33: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/33.jpg)
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
![Page 34: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/34.jpg)
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%?
![Page 35: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/35.jpg)
Remember grain is made up of dry matter + H2O
The amount of H2O changes, but the amount of dry matter in bu is constant.
![Page 36: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/36.jpg)
Standard bu
m
t
m
W
W
W0.135
![Page 37: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/37.jpg)
51.9W
W0.25
m
m
![Page 38: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/38.jpg)
So water removed =H2O @ 25% - H2O @ 13.5%
![Page 39: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/39.jpg)
Your turn: How much water needs to be removed
to dry shelled corn from 23% (wb) to 15% (wb) if we have 1000 bu?
![Page 40: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/40.jpg)
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
![Page 41: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/41.jpg)
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
![Page 42: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/42.jpg)
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
![Page 43: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/43.jpg)
Enthalpy = ____ BTU/lbda
Humidity ratio=______ lbH2O/lbda
Specific volume = ______ ft3/lbda
Dew point temp = _____ F
![Page 44: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/44.jpg)
Psychrometric Processes Sensible heating – horizontally to the
right Sensible cooling – horizontally to the left
Note that RH changes without changing the humidity ratio
![Page 45: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/45.jpg)
Psychrometric Processes Evaporative cooling = grain drying (p
266)
![Page 46: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/46.jpg)
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?
![Page 47: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/47.jpg)
Solution
@ 24C, 68% RH: Enthalpy = 56 kJ/kgda
@ 46C: Enthalpy = 78 kJ/kgda
V = 0.922 m3/kgda
![Page 48: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/48.jpg)
V
ΔhQEnergy
kg
kJ22Δh
da
![Page 49: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/49.jpg)
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
![Page 50: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/50.jpg)
![Page 51: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/51.jpg)
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
![Page 52: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/52.jpg)
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)
![Page 53: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/53.jpg)
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
![Page 54: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/54.jpg)
Sample EMC
![Page 55: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/55.jpg)
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?
![Page 56: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/56.jpg)
Drying Calculations
![Page 57: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/57.jpg)
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.
![Page 58: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/58.jpg)
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
![Page 59: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/59.jpg)
Step 1
How much water must be removed?
2000 bu
20% to 13%
Now what?
![Page 60: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/60.jpg)
Step 1Std bu = 60 lb @ 0.135mw =
md = mt – mw =
![Page 61: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/61.jpg)
Step 1
![Page 62: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/62.jpg)
Step 2
How much water can each pound of dry air remove?
How do we approach this step?
![Page 63: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/63.jpg)
Step 2Find exit conditions from EMC.Plot on psych chart.
0C = 32F = 64%10C = 50F = 67%30C = 86F = 72%
![Page 64: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/64.jpg)
Step 2
@ 52F – 68% RH
![Page 65: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/65.jpg)
Change in humidity ratio
![Page 66: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/66.jpg)
Each pound of dry air can remove
![Page 67: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/67.jpg)
We need to remove 10,500 lbH2O.
Each lbda removes 0.0023 lbH2O.
![Page 68: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/68.jpg)
Step 3Determine the cubic feet of air we need to remove necessary water
![Page 69: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/69.jpg)
Step 3 Calculations
![Page 70: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/70.jpg)
Step 4Determine the fan operating speed
How do we approach this?
![Page 71: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/71.jpg)
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
![Page 72: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/72.jpg)
Step 4
½
Fgrain
6300 cfmQ
PS
![Page 73: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/73.jpg)
From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.
![Page 74: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/74.jpg)
![Page 75: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/75.jpg)
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.
![Page 76: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/76.jpg)
Example 2With heated air, is conserved (not Q)
m
2ptQ
m
![Page 77: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/77.jpg)
Example 2
2. Determine the relative humidity of the air leaving the drier.
![Page 78: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/78.jpg)
Example 2
32 40.5 118
78% RH
![Page 79: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/79.jpg)
Example 2
3. Determine the amount of propane fuel required per hour.
![Page 80: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/80.jpg)
Example 2
mΔh
h
h
kg
kJ50,000Propane
1
2
fuel
![Page 81: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/81.jpg)
Example 2
4. Determine the amount of fruit residue dried per hour.
![Page 82: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/82.jpg)
Example 2
@ 85%, 0.15 of every kg is dry matter
![Page 83: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/83.jpg)
Example 2
Remove 0.85 – 0.0423 = wetresidue
OH
kg
kg0.8077 2
ΔH
![Page 84: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/84.jpg)
Example 2
![Page 85: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/85.jpg)
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 -
![Page 86: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/86.jpg)
1. What is the necessary fan delivery rate (cfm)?
![Page 87: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/87.jpg)
2. What is the approximate total pressure drop (in inches of water) required to obtain the needed air flow?
![Page 88: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/88.jpg)
3. The estimated fan HP based on fan efficiency of 65%
![Page 89: Bernoulli’s Theorem for Fans](https://reader035.vdocuments.us/reader035/viewer/2022081503/56815acf550346895dc8a280/html5/thumbnails/89.jpg)
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.