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
Bernoulli’s Theorem for FansPE Review Session VIB – section 1
Fan and Bin
γ
Ph
γ
Ph
2g
v
γ
PhFW
2g
v
γ
Ph
33
11
233
3
211
1
12
3
Fγ
P
2g
v
γ
PW
FW
0vv
T222
31
staticpressure
velocityhead
total pressure
Power
s
s
T
T
e
QPP
or
e
QPP
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)
ASABE Standards - graph for Ffloor
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
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 m2. Find the static and total pressure when Q=4 m3/s
F=F(pipe)+F(exp)+F(floor)+F(grain) F(pipe)=
g2
v
D
LfF
2
pipe
s
msm
4.20
45.0
4
A
QV
2
3
pipepipe
f
Dv
andD
Re
Re
5
5
3
107.6Re
1082.1
202.14.205.0Re
sPamkg
sm
m
)(
103
5.01000
115.0
4
moodyf
mmmm
mm
D
m
smsm
m
mF
f
pipe
2.3
81.92
4.20
5.0
5015.0
015.0
2
2
Fexp
g
vvF
2
22
21
exp
m
sm
sm
F 2.2181.92
04.20
2
2
exp
Ffloor Equ. 2.39
g
ov
msPa
pf
2
2
2
071.1
V = Vbin = s
m
msm
A
Q
bin
2.020
4
2
3
Of=0.1
4.0p
m
sm
mkgmsPa
Ffloor 3.281.9202.1
4.01.02.0
071.1
22
2
2
Fgrain
Pa
sm
msm
P
bV
cVa
L
PF fwheat
1599
2.077.81ln
15.12.0107.2
1ln2
4
2
1599 Pa = _________ m?
m
sm
mkg
mN
gmN
13581.9202.1
15991599
23
22
Using Shedd’s CurvesV=0.2 m/sWheat
mPacmm
Pa
L
Pf 127150011000
Ftotal = 3.2 + 21.2 + 2.3 + 130
= 157 m
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 and PsychrometricsCore 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)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.
Standard bu
51.9lb8.1lb60lbW
8.1lb)0.135(60lbW
60lb
W
W
W0.135
d
m
m
t
m
17.3lb0.75
13W
0.75W13
W130.25W
51.9W
W0.25
m
m
mm
m
m
So water removed =H2O @ 25% - H2O @ 13.5%
O18,400lbH2000bu*bu
lb9.2
bu
lb9.2
bu
lb8.1
bu
lb17.3
2
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
Psychrometric Processes Evaporative cooling = grain drying (p
266)
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?
Solution
@ 24C, 68% RH: Enthalpy = 56 kJ/kgda
@ 46C: Enthalpy = 78 kJ/kgda
V = 0.922 m3/kgda
119kW
60s
1min
min
300m
kgm
0.922
kgkJ
22
V
ΔhQEnergy
kg
kJ22Δh
3
da
3da
da
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
Sample EMC
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?
Drying Calculations
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 1
How much water must be removed?
2000 bu
20% to 13%
Now what?
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 1
OH
2
w
w
w
210440lb2000bu
bu
lb5.22
bu
lb5.227.76lb12.98lb
:OΔH
12.98lbm
51.9lbm
m0.2
Step 2
How much water can each pound of dry air remove?
How do we approach this step?
Step 2Find exit conditions from EMC.Plot on psych chart.
0C = 32F = 64%10C = 50F = 67%30C = 86F = 72%
Step 2
@ 52F – 68% RH
Change in humidity ratio
Each pound of dry air can remove
da
OH
da
OH
lb
lb0.0023
lb
lb0.00330.0056 22
We need to remove 10,500 lbH2O.
Each lbda removes 0.0023 lbH2O.
OH
daOHda
2
2 0.0023lb
1lb10500lbb4,565,217l
Step 3Determine the cubic feet of air we need to remove necessary water
Step 3 Calculations
da
3
da3air lb
ft13.2b4,565,217lft60,260,870
Step 4Determine the fan operating speed
How do we approach this?
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
Step 4
½
Fgrain
6300 cfmQ
PS
From cfm of fan and cubic feet of air, determine the time needed to dry the soybeans.
6.6d
159hrs
9565min
minft
6300
ft60,260,8703
3
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
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.
Example 2
hr
kg44
s
kJ615.4mΔh
kg
kJ49.5h
kg
kJ140h
kg
kJ50,000Propane
propane
da1
da2
fuel
Example 2
4. Determine the amount of fruit residue dried per hour.
Example 2
@ 85%, 0.15 of every kg is dry matter
OHm
m
m
20.0423kgw
0.15w
w0.22
Example 2
Remove 0.85 – 0.0423 = wetresidue
OH
kg
kg0.8077 2
da
OH
kg
kg0.0320.0060.038ΔH 2
Example 2
hr
kg970
0.8077kg
kg
s
6.8kg
kg
kg0.032
wetfruit
OH
wetresidueda
da
OH
2
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 -
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.