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7/27/2019 2011 Core Ag Engineering Principles - Session II
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Pump Affinity Laws
7/27/2019 2011 Core Ag Engineering Principles - Session II
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Pump Affinity LawsP. 100 of text
–section 4: vary
only speed of pump
P. 100 of text – section 5:vary only diameter
P. 106 of text – vary BOTHspeed and diameter ofimpeller.
3
2
1
2
1
2
1)(
D
D
N
N
Q
Q
2
2
1
2
2
1
2
1
D
D
N
N
W
W
7/27/2019 2011 Core Ag Engineering Principles - Session II
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Power out equations
2
1
5
2
1
3
2
1
2
1
D
D
N
N
P
P
o
o
g QW P o
p. 106
p. 89
7/27/2019 2011 Core Ag Engineering Principles - Session II
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A pump is to be selected that is geometricallysimilar to the pump given in the performancecurve below, and the same system. What D
and N would give 0.005 m3
/s against a headof 19.8 m?
900W 9m
1400W
W
0.01 m3 /s
D = 17.8 cm
N = 1760 rpm
7/27/2019 2011 Core Ag Engineering Principles - Session II
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What is the operating point of first pump?
N1 = 1760
D1 = 17.8 cmQ1 = 0.01 m3/s Q2 = 0.005 m3/s
W1 = 9m W2 = 19.8 m
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Now we need to “map” to
new pump on same systemcurve.
Substitute into Solve for D2
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7/27/2019 2011 Core Ag Engineering Principles - Session II
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N2
= ?
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Try it yourself
If the system used in the previousexample was changed by removing alength of pipe and an elbow – what
changes would that require you tomake?
Would N1 change? D1? Q1? W1? P1?
Which direction (greater or smaller)would “they” move if they change?
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Moisture and PsychrometricsCore Ag Eng Principles Session IIB
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Moisture in biological products
can be expressed on a wet basisor dry basis
wet basis
dry basis (page 273)
d
m
dm
m
W
WM
)W(W
Wm
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Standard bushels
ASAE Standards
Corn weighs 56 lb/bu at 15% moisturewet-basis
Soybeans weigh 60 lb/bu at 13.5%moisture wet-basis
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Use this information todetermine how much water needs to be removed to dry
grain We have 2000 bu of soybeans at 25%moisture (wb). How much water mustbe removed to store the beans at
13.5%?
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Remember grain is made up of drymatter + H2O
The amount of H2O changes, but the
amount of dry matter in bu is constant.
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Standard bu
51.9l8.1lb60lbW
8.1lb)0.135(60lbW
60lb
W
W
W0.135
d
m
m
t
m
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17.3l0.7513W
0.75W13
W130.25W
51.9W
W0.25
m
m
mm
m
m
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So water removed =H2O @ 25% - H2O @ 13.5%
O18,400lbH2000bu*bu
lb9.2
bu
lb9.2bu
lb8.1bu
lb17.3
2
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Your turn:
How much water needs to be removedto dry shelled corn from 23% (wb) to15% (wb) if we have 1000 bu?
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Psychrometrics
If you know two properties of anair/water vapor mixture you know allvalues because two propertiesestablish a unique point on the psychchart
Vertical lines are dry-bulb temperature
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Psychrometrics
Horizontal lines are humidity ratio(right axis) or dew point temp (leftaxis)
Slanted lines are wet-bulb temp andenthalpy
Specific volume are the “other” slantedlines
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Your turn:
List the enthalpy, humidity ratio,specific volume and dew pointtemperature for a dry bulb
temperature of 70F and a wet-bulbtemp of 60F
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Enthalpy = 26 BTU/lbda Humidity ratio=0.0088 lbH2O/lbda
Specific volume = 13.55 ft3/lbda
Dew point temp = 54 F
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Psychrometric Processes
Sensible heating – horizontally to theright
Sensible cooling – horizontally to the
left
Note that RH changes without
changing the humidity ratio
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Example
A grain dryer requires 300 m3/min of 46C air. The atmospheric air is at 24Cand 68% RH. How much power must
be supplied to heat the air?
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Solution
@ 24C, 68% RH: Enthalpy = 56 kJ/kgda
@ 46C: Enthalpy = 78 kJ/kgda
V = 0.922 m3/kgda
7/27/2019 2011 Core Ag Engineering Principles - Session II
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119kW
60s
1min
min
300m
kg
m0.922
kgkJ22
V
ΔhQEnergy
kg
kJ22 Δh
3
da
3
da
da
7/27/2019 2011 Core Ag Engineering Principles - Session II
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Equilibrium Moisture Curves
When a biological product is in a moistenvironment it will exchange water with theatmosphere in a predictable way – depending on the temperature/RH of themoist air surrounding the biological product.
This information is contained in the EMC for each product
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7/27/2019 2011 Core Ag Engineering Principles - Session II
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Equilibrium Moisture Curves
Establish second point on theevaporative cooling line – i.e. can’t
remove enough water from the
product to saturate the air under allconditions – sometimes the exhaustair is at a lower RH because the
product won’t “release” any morewater
7/27/2019 2011 Core Ag Engineering Principles - Session II
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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 four T/RH points from EMCs
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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 thestate of the exhaust air
7/27/2019 2011 Core Ag Engineering Principles - Session II
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Sample EMC
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We are drying corn to 15% wb;with natural ventilation usingoutside 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
5140-9000 cfm at ½” H2O staticpressure. The bin is 26’ in diameter
and outside air (60 F, 30% RH) is
being blown over the soybeans.
7/27/2019 2011 Core Ag Engineering Principles - Session II
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Steps to work drying problem
Determine how much water needs to beremoved (from moisture content before andafter; 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 dryproduct
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Step 1
How much water must beremoved?
2000 bu
20% to 13%
Now what?
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Step 1
Std 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
6.70.13mm
51.9m
m0.13
w
ww
w
w
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Step 2
How much water can eachpound 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 2
@ 52F –
68% RH
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Change in humidity ratio
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7/27/2019 2011 Core Ag Engineering Principles - Session II
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We need to remove 10500lbH2O.
Each lbda removes 0.0023
lbH2O.
OH
daOHda
2
2 0.0023lb1lb10500lbb4,565,217l
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Step 3 Calculations
da
3
da3air
lb
ft13.2b4,565,217lft60,260,870
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Step 4
Determine the fan operatingspeed
How do we approach this?
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Step 4
Main term in F is Fgrain
Airflow (cfm/ft2)
5030
15
10
Pressure drop (“H2O/ft)
0.50.23
0.09
0.05
x depth x CF
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Step 4
½
Fgrain
6300 cfmQ
PS
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6.6d
159hrs
9565mi
min
ft6300
ft60,260,8703
3
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Example 2
Ambient air at 32C and 20% RH isheated to 118 C in a fruit residuedryer. The flow of ambient air into
the propane heater is at 5.95m3/sec. The drying is to be carriedout from 85% to 22% wb. The air leaves the drier at 40.5C.
Determine the airflow rate of theheated air.
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Example 2
2. Determine the relative humidityof the air leaving the drier.
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Example 2
32 40.5 118
78% RH
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Example 2
3. Determine the amount of propane fuel required per hour.
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Example 2
4. Determine the amount of fruitresidue dried per hour.
E l 2
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Example 2
@ 85%, 0.15 of every kg is drymatter
OHm
m
m
2
0.0423kgw
0.15w
w0.22
E l 2
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Example 2
Remove 0.85 – 0.0423 = wetresidue
OH
kg
kg0.8077 2
da
OH
kg
kg0.0320.0060.038 ΔH
2
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Example 2
hr kg970
0.8077kg
kg
s
6.8kg
kg
kg0.032
wetfruit
OH
wetresidueda
da
OH
2
2
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Your Turn:
A grain bin 26’ in diameter has aperforated floor over a plenumchamber.
Shelled field corn will be dried from an
initial mc of 24% to 14% (wb). Batch
drying (1800 std. bu/batch) will beused
with outside air (55F, RH 70%) that hasbeen heated 10F before being passed
through the corn. To dry the corn in 1
week -
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1. What is the necessary fandelivery rate (cfm)?
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2. What is the approximate totalpressure drop (in inches of water)required to obtain the needed air
flow?
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3. The estimated fan HP based onfan efficiency of 65%
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