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P.E. Review Session

V–C. Mass Transfer between Phases

by
USDA-ARS
Center for Grain and Animal Health Research
Manhattan, Kansas
Current NCEES Topics
Primary coverage:                        Exam %
   V. C. Mass transfer between phases        4%
 I. D. 1. Mass and energy balances           ~2%
Also:
 I. B. 1. Codes, regs., and standards        1%
Overlaps with:
 I. D. 2. Applied psychrometric processes    ~2%
 II. A. Environment (Facility Engr.)         3-4%
Specific Topics/Unit Operations
   Heat & mass balance fundamentals
   Evaporation (jam production)
   Postharvest cooling (apple storage)
   Sterilization (food processing)
   Heat exchangers (food cooling)
   Drying (grain)
   Evaporation (juice)
   Postharvest cooling (grain)
Mass Transfer between Phases

   A subcategory of: Unit Operations
   Common operations that constitute a process, e.g.:
   pumping, cooling, dehydration (drying), distillation,
evaporation, extraction, filtration, heating, size reduction,
and separation.

   How do you decide what unit operations apply to
a particular problem?
   Experience is required (practice).
Principles

   Mass Balance
Inflow = outflow + accumulation

   Energy Balance
Energy in = energy out + accumulation

   Specific equations
Fluid mechanics, pumping, fans, heat transfer,
drying, separation, etc.
Illustration – Jam Production

Jam is being manufactured from crushed fruit with
14% soluble solids.
   Sugar is added at a ratio of 55:45
   Pectin is added at the rate of 4 oz/100 lb sugar
The mixture is evaporated to 67% soluble solids
What is the yield (lbjam/lbfruit) of jam?
Illustration – Jam Production
mv = ?

mf = 1 lbfruit (14% solids)

ms = 1.22 lbsugar

mp = 0.0025 lbpectin                   mJ = ? (67% solids)
Illustration – Jam Production
mv = ?

mf = 1 lbfruit (14% solids)

ms = 1.22 lbsugar

m = 0.0025 lb
p      pectin                        mJ = ? (67% solids)
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
Illustration – Jam Production
mv = ?

mf = 1 lbfruit (14% solids)

ms = 1.22 lbsugar

mp = 0.0025 lbpectin                          mJ = ? (67% solids)
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0

Solids Balance:
Inflow = Outflow + Accumulation
mf·Csf   + ms·Css = mJ·CsJ + 0.0
(1 lb)·(0.14lb/lb) + (1.22 lb)·(1.0lb/lb) = mJ·(0.67lb/lb)
Illustration – Jam Production
mv = ?

mf = 1 lbfruit (14% solids)

ms = 1.22 lbsugar

mp = 0.0025 lbpectin                          mJ = ? (67% solids)
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0

Solids Balance:
Inflow = Outflow + Accumulation
mf·Csf   + ms·Css = mJ·CsJ + 0.0
(1 lb)·(0.14lb/lb) + (1.22 lb)·(1.0lb/lb) = mJ·(0.67lb/lb)

mJ = 2.03 lbJam/lbfruit

mv = 0.19 lbwater/lbfruit
Illustration – Jam Production
mv = ?

mf = 1 lbfruit (14% solids)

ms = 1.22 lbsugar

mp = 0.0025 lbpectin                                mJ = ? (67% solids)

What if this was a continuous flow concentrator
with a flow rate of 10,000 lbfruit/h?
Principles

• Mass Balance:
Inflow = outflow + accumulation                    Ci
Chemical                    
m1                       t
concentrations:
Ci ,1

m2    Ci,2
• Energy Balance:
Energy in = energy out + accumulation
m  mass flow rate, kg/s
                                                             T

m1
T  temperature, K                                             t
c p  specific heat capacity, J/kg  K       T1        
m2      T2
Principles

• Mass Balance:
Inflow = outflow + accumulation
Chemical                                               Ci
concentrations:      Ci ,1  m1  Ci , 2  m2   V 
             
t

• Energy Balance:
Energy in = energy out + accumulation
T
m1  c p  T1  m2  c p  T2    c p V 
               
t
(sensible energy) total energy = m·h
Illustration − Apple Cooling

An apple orchard produces 30,000 bu of apples a year, and
will store ⅔ of the crop in refrigerated storage at 31°F. Cool
to 34°F in 5 d; 31°F by 10 d.
…
Estimate the refrigeration requirements for the 1st 30 days.
Apple Cooling
qfrig
Principles

   Mass Balance
Inflow = outflow + accumulation

   Energy Balance
Energy in = energy out + accumulation

   Specific equations
Fluid mechanics, pumping, fans, heat transfer,
drying, separation, etc.
Illustration − Apple Cooling
energy in = energy out + accumulation
qfrig
qin,1+ ... = qout,1+ ... + qa
Illustration − Apple Cooling

Try it...
Illustration − Apple Cooling

Try it...

An apple orchard produces 30,000 bu of apples a year, and
will store ⅔ of the crop in refrigerated storage at 31°F. Cool
to 34°F in 5 d; 31°F by 10 d.
…
Estimate the refrigeration requirements for the 1st 30 days.
Apple Cooling
qfrig

qm

qso      qr                  qb
qe
qs   qm
qin
Apple Cooling
   Sensible heat terms…
qs = sensible heat gain from apples, W
qr = respiration heat gain from apples, W
qm = heat from lights, motors, people, etc., W
qso = solar heat gain through windows, W
qb = building heat gain through walls, etc., W
qin = net heat gain from infiltration, W
qe = sensible heat used to evaporate water, W

1 W = 3.413 Btu/h, 1 kW = 3413. Btu/h
Apple Cooling
   Sensible heat equations…
qr = mtot· Hresp
qm = qm1 + qm2 + . . .
qb = Σ(A/RT)· (Ti – To)
0
qin = (Qacpa/vsp)· (Ti – To)
0
qso = ...
Apple Cooling
   definitions…
Hresp = sp. rate of heat of respiration, J/kg·s (Btu/lb·h)
mtot = total mass of apples, kg (lb)
cpA = sp. heat capacity of apples, J/kg·°C (Btu/lb°F)
cpa = specific heat capacity of air, J/kg·°C (Btu/lb°F)
Qa = volume flow rate of infiltration air, m3/s (cfm)
vsp = specific volume of air, m3/kgDA (ft3/lbDA)
A = surface area of walls, etc., m2 (ft2)
RT = total R-value of walls, etc., m2·°C/W (h·ft2·°F/Btu)
Ti = air temperature inside, °C (°F)
To = ambient air temperature, °C (°F)
qm1, qm2 = individual mechanical heat loads, W (Btu/h)
Example 1

An apple orchard produces 30,000 bu of apples a year, and will
store ⅔ of the crop in refrigerated storage at 31°F. Cool to 34°F
in 5 day; 31°F by 10 day.
declines to 65°F in 20 days
A = 46 lb/bu; cpA = 0.9 Btu/lb°F

What is the sensible heat load from the apples on day 3?
Example 1
qfrig

qm

qso    qr                  qb
qe
qs   qm
qin
Example 1
mload = (2000 bu/day · 3 day)·(46 lb/bu)
mload = 276,000 lb   (on day 3)

ΔT = (75°F – 34°F)/(5 day) = 8.2°F/day

qs = (276,000 lb)·(0.9 Btu/lb°F)·(8.2°F/day)
qs = 2,036,880 Btu/day = 7.1 ton
(12,000 Btu/h = 1 ton refrig.)
Example 1, revisited
mload = 276,000 lb   (on day 3)
Ti,avg = (75 + 74.5 + 74)/3 = 74.5°F

ΔT = (74.5°F – 34°F)/(5 day) = 8.1°F/day

qs = (276,000 lb)·(0.9 Btu/lb°F)·(8.1°F/day)
qs = 2,012,040 Btu/day = 7.0 ton
(12,000 Btu/h = 1 ton refrig.)
Example 2

Given the apple storage data of example 1,
 = 46 lb/bu; cpA = 0.9 Btu/lb°F; H = 3.4 Btu/lb·day

What is the respiration heat load (sensible) from the
apples on day 1?
Example 2

qr = mtot· Hresp

mtot = (2000 bu/day · 1 day)·(46 lb/bu)
mtot = 92,000 lb

qr = (92,000 lb)·(3.4 Btu/lb·day)
qr = 312,800 Btu/day = 1.1 ton
   Sterilization
   Heat exchangers
   Drying
   Evaporation
   Postharvest cooling
Sterilization

   First order thermal death rate (kinetics) of
microbes assumed (exponential decay)
N  No ek   D   t

   D = decimal reduction time = time, at a given
temperature, in which the number of microbes is
reduced 90% (1 log cycle)
 N               t
ln 
N    kD  t 
 o               D
Sterilization
( 250 FT )

Thermal death time: t  Fo 10                 z
   The z value is the temperature increase that will result in a
tenfold increase in death rate
   The typical z value is 10°C (18°F) (C. botulinum)
   Fo = time in minutes at 250°F that will produce the same degree
of sterilization as the given process at temperature T
   Standard process temp = 250°F (121.1°C)
   Thermal death time: given as a multiple of D
   Pasteurization: 4 − 6D
 Milk: 30 min at 62.8°C (“holder” method; old batch method)
15 sec at 71.7°C (HTST − high temp./short time)
   Sterilization: 12D
   “Overkill”: 18D (baby food)
Sterilization
z

Thermal Death Time Curve
(C. botulinum)
(Esty & Meyer, 1922)
( 250 FT )

t  Fo 10        z

t = thermal death time, min
z = DT for 10x change in t, °F
2.7
Fo = t @ 250°F (std. temp.)
Sterilization
10

Thermal Death Rate Plot
(Stumbo, 1949, 1953; ...)                                                  z

Decimal Reduction Time
N  No ek   D   t                                    1

D = decimal reduction time

Dr = 0.2
 N  t
N  D
ln                                                 0.1
 o

0.01
121.1
100     110      120       130

Temperature, °C
Sterilization equations
             
 ( 250  T )                      D To  T

             
                 log    
DT  D250 10      z                            Do   z

No                                N o  Fo FT
Fo  D250   log                              log        
N                                 N  Do DT

                                          
 (T  250 F )              (T 121.1C ) 

               
            
               

Fo  t 10         z       
Fo  t 10
      z        
Sterilization

   Popular problems would be:
−   Find a new D given change in temperature
−   Given one time-temperature sterilization process,
find the new time given another temperature, or
the new temperature given another time
Example 3
   If D = 0.25 min at 121°C, find D at 140°C.
z = 10°C.
Example 3

equation     log
D To  T

D121 = 0.25 min
Do   z       z = 10°C
substitute         D140     121.1C  140C
log          
0.25 min        10C

solve        ...

Example 4

   The Fo for a process is 2.7 minutes. What
would be the processing time if the processing
temperature was changed to 100°C?

NOTE: when only Fo is given, assume standard
processing conditions:
T = 250°F (121.1°C); z = 18°F (10°C)
Example 4

Thermal Death Time Curve
(C. botulinum)
(Esty & Meyer, 1922)
(121.1 C T )

t  Fo 10         z

t = thermal death time, min
z = DT for 10x change in t, °C
2.7
Fo = t @ 121.1°C (std. temp.)
Example 4
(121.1 C T )
t  Fo 10         z

(121.1 C 100 C )
t100  (2.7 min) 10         10 C

t100  348 min
Heat Exchanger Basics

q  U  Ae  DTm  U  A  DTlm

 DT  DT          (T  T )  (T  T )                (T  T )  (T  T ) 
DTlm     max DT min        Hi  Co         Ho Ci 
  Hi    Ci         Ho Co 

 
 ln DTmax
       min



          
T T
ln THi  TCo
Ho

Ci

 counter

           
THi  T
ln THo  TCi
Co
    
 parallel

mH cH DTH  mC cC DTC  q
           
Heat Exchangers

subscripts:   H – hot fluid    i   – side where the fluid enters
C – cold fluid   o   – side where the fluid exits
variables: m = mass flow rate of fluid, kg/s
c = cp = heat capacity of fluid, J/kg-K
C = mc, J/s-K
U = overall heat transfer coefficient, W/m2-K
A = effective surface area, m2
DTm = proper mean temperature difference, K or °C
q = heat transfer rate, W
F(Y,Z) = correction factor, dimensionless
Example 5

   A liquid food (cp = 4 kJ/kg°C) flows in the inner pipe of
a double-pipe heat exchanger. The food enters the
heat exchanger at 20°C and exits at 60°C. The flow
rate of the liquid food is 0.5 kg/s. In the annular
section, hot water at 90°C enters the heat exchanger
in counter-flow at a flow rate of 1 kg/s. Assuming
steady-state conditions, calculate the exit temperature
of the water. The average cp of water is 4.2 kJ/kg°C.
Example 5
90°C

   Solution
60°C   ?
mf cf DTf = mw cw DTw
20°C
(0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C)
= (1 kg/s)·(4.2 kJ/kg°C)·(90 – THo)

THo = 71°C
Example 6
   Find the heat exchanger area needed from
example 5 if the overall heat transfer coefficient
is 2000 W/m2·°C.
Example 6
   Find the heat exchanger area needed from
example 5 if the overall heat transfer coefficient
is 2000 W/m2·°C.
Data:
liquid food, cp = 4 kJ/kg°C
water, cp = 4.2 kJ/kg°C
Tfood,inlet = 20°C, Tfood,exit = 60°C
Twater,inlet = 90°C
mfood = 0.5 kg/s
mwater = 1 kg/s
Example 6
DTmin = 90°–60°C
90°C

DTmax = 71°–20°C
   Solution                                                      71°C
60°C

q  U  Ae  DTlm   q  mC cC DTC

20°C

q = mf cf DTf = (0.5 kg/s)·(4 kJ/kg°C)·(60 – 20°C) = 80 kJ/s
DTlm = (DTmax – DTmin)/ln(DTmax/DTmin) = 39.6°C
Ae = (80 kJ/s)/{(2 kJ/s·m2·°C)·(39.5°C)}
2000 W/m2·°C = 2 kJ/s·m2·°C
Ae = 1.01 m2

   Effectiveness ratio (H, P, & Young, pp. 204-212)
(Ta1  Ta 2 )             UA             Cb
Ecooling                    ,   NTU        ,      R
(Ta1  Tb ,in )           Cm in           Ca

   One fluid at constant T: R
   DTlm correction factors
q  U  A  DTlm  F ( Z , Y )
Time Out
Reference Ideas
Need                          Mark’s Suggestion
   Full handbook      The one you use regularly
 ASHRAE Fundamentals.

   Processing text    Henderson, Perry, & Young (1997),
Principles of Processing Engineering
 Geankoplis (1993), Transport Processes
& Unit Operations.
   Standards            ASABE Standards, recent ed.
   Other text         Albright (1991), Environmental Control...
 Lower et al. (1994), On-Farm Drying and...
 MWPS-29 (1999), Dry Grain Aeration
Systems Design Handbook. Ames, IA: MWPS.
Studying for & taking the exam

   Practice the kind of problems you plan to
work
   Know where to find the data
   See presentation I-C Economics and Statistics,
on Preparing for the Exam
Mass Transfer Between Phases
   Psychrometrics
   A few equations
   Psychrometric charts
(SI and English units, high, low and normal temperatures; charts
in ASABE Standards)

   Psychrometric Processes – Basic Components:
   Sensible heating and cooling
   Humidify or de-humidify
   Drying/evaporative cooling
Mass Transfer Between Phases
cont.

   Grain and food drying                       Twb

   Sensible heat
Psychrometrics
   Latent heat of vaporization
   Moisture content: wet and dry basis, and equilibrium
moisture content (ASAE Standard D245.6)
   Airflow resistance (ASAE Standard D272.3)
Mass Transfer Between Phases
cont.

25
Effect of temperature on
20       moisture isotherms (corn data)
Equilibrium Moisture
Content, %

15

10                                                0°C
20°C
5                                                40°C

0
0         20        40         60       80          100
Relative Humidity, %
Mass Transfer Between Phases
cont.

25
Equilibrium Moisture Content, %

20
ASAE Standard D245.6 –                                                         .

15    Use previous revision (D245.4) for constants
10                 or                      0°C
20°C
use psychrometric charts in Loewer et al.   (1994)
40°C
5

0
0   20       40         60         80        100
Relative Humidity, %
Mass Transfer Between Phases
cont.

Loewer, et al. (1994)
Mass Transfer Between Phases
cont.

25
Effect of temperature on
20       moisture isotherms (corn data)
Equilibrium Moisture
Content, %

15

10                                                0°C
20°C
5                                                40°C

0
0         20        40         60       80          100
Relative Humidity, %
Deep Bed Drying Process

rhe

Twb
rho

TG   To
Use of Moisture Isotherms

Air Temp.
Grain Temp.
Equilibrium Moisture Content, %

Mo

To              TG
Me

rho                              rhe
Relative Humidity, %
Drying
Deep Bed

 Drying grain (e.g., shelled corn) with the drying air
flowing through more than two to three layers of
kernels.
 Dehydration of solid food materials

   ≈ multiple layers drying & interacting
(single, thin-layer solution is a single equation)
1                           1
M wb                   M db     
1  M db                    1  M wb
W1  (1  M wb ,1 )  W2  (1  M wb ,2 )
Drying
Deep Bed vs. Thin Layer

   Thin-layer process is not as complex. The common
 k t n
Page eqn. is: MR  e           (falling rate drying period)
   Definitions:
k, n = empirical constants (ANSI/ASAE S448.1)
t = time
M  M equilibrium
MR                            ; M  dry basis moisture content
M initial  M equilibrium
   Deep bed effects when air flows through more than two
to three layers of kernels.
Grain Bulk Density
for deep bed drying calculations

kg/m3               lb/bu[1]
Corn, shelled                  721                  56
Milo (sorghum)                 721                  56
Rice, rough                    579                  45
Soybean                        772                  60
Wheat                          772                  60
1Standard   bushel.   Source: ASAE D241.4
Basic Drying Process
Mass Conservation

   Compare:        moisture added to air
to
moisture removed from product
Basic Drying Process
Mass Conservation

humidity ratio : a,out

Da  a ,out  a ,in      mg  total mass of grain

DWg  change in grain MC

ma
Fan
humidity ratio : a,in
Basic Drying Process
Mass Conservation

Try it:

   Total moisture conservation equation:
Basic Drying Process
Mass Conservation

   Compare:        moisture added to air
to
moisture removed from product
   Total moisture conservation:
kga  kgw                    kgw
kgg
s s kga                    kgg

ma  t  Da  mg  DWg

Basic Drying Process
Mass Conservation – cont’d

   Calculate time:
mg DWg
t
ma Da


   Assumes constant outlet conditions (true initially)
   but outlet conditions often change as product dries…
   use “deep-bed” drying analysis for non-constant outlet
conditions
(Henderson, Perry, & Young sec. 10.6 for complete analysis)
Drying Process
time varying process

Drying Rate
(Thin-layer)

Constant
Falling

Rate
Rate

Time     →
Evaporative    erh = 100%                erh < 100%
Cooling        aw = 1.0                  aw < 1.0

 Assume falling rate period, unless…
 Falling rate requires erh or exit air data
Drying Process
cont.

erh
ASAE D245.6
Twb
Example 7

   Hard wheat at 75°F is being dried from 18% to 12%
w.b. in a batch grain drier. Drying will be stopped
when the top layer reaches 13%. Ambient
conditions: Tdb = 70°F, rh = 20%
   Determine the exit air temperature early in the
drying period.
   Determine the exit air RH and temperature at the
end of the drying period?
Example 7

 Part II
Use Loewer, et al. (1994 ) (or ASAE D245.6)

RHexit = 55%            rhexit                 emc=13%

Twb
Texit = 58°F

Texit
Example 7
13%

Loewer, et al. (1994)
Example 7b

 Part I
Use Loewer, et al. (1994 ) (or ASAE D245.6)

Texit = Tdb,e = TG                            emc=18%

Twb

Tdb,e
Example 7b
18%

53.5

Loewer, et al. (1994)
Example 7b

 Part I
Use Loewer, et al. (1994 ) (or ASAE D245.6)

Texit = Tdb,e = TG = 53.5°F                   emc=18%

Twb

Tdb,e
Cooling Process
Energy Conservation

   Compare:          heat added to air
to
heat removed from product
   Sensible energy conservation:
ma t ca DTa  mg c g DTg


   Total energy conservation:              DTg  Tinitial  TII

ma t Dha  mg c g DTg

Cooling Process
(and Drying)

erh

Twb
Airflow in Packed Beds
Drying, Cooling, etc.
Design Values for Airflow Resistance in Grain
100

Soybeans (PF=1.3)

Corn (PF=1.5)
10
2
Airflow, cfm/ft

Sunflower (PF=1.5)                       Barley (PF=1.5)
Milo (PF=1.3)
1                                     Wheat (PF=1.3)

0.1
0.001             0.01                 0.1                  1   10
Pressure Drop per Foot, inH2O/ft
Source: ASABE D272.3, MWPS-29
Aeration Fan Selection

Pressure drop (loose fill, “Shedd’s data”):
DP = (inH2O/ft)LF x MS x (depth) + 0.5
Shedd’s curve multiplier
(Ms = PF = 1.3 to 1.5)
Pressure drop (design value chart):
DP = (inH2O/ft)design x (depth) + 0.5
Aeration Fan Selection

Pressure drop (loose fill, “Shedd’s data”):
DP = (inH2O/ft)LF x MS x (depth) + 0.5

Pressure drop (design value chart):
DP = (inH2O/ft)design x (depth) + 0.5

0.5 inH2O pressure drop in ducts -
Standard design assumption
(neglect for full perforated floor)
Standards, Codes, & Regulations

   Standards
   ASABE
   Already mentioned ASAE D245.6 and D272.3
   ASAE D243.3 Thermal properties of grain and…
   ASAE S448 Thin-layer drying of grains and crops
   Several others
   Others not likely for unit operations
More Examples
Evaporator (Concentrator)
mV

mF        Juice
mP

mS
Evaporator
   Solids mass balance:
mF  X F mP  X P
                           X  Concentration, lb
lb

   Total mass balance:
mF  mV mP
       
   Total energy balance:

mF  c pF  TF  mS  (h fg ) S  mV  hgv mP  c pP  TP
                                         
Example 8

Fruit juice concentrator, operating @ T =120°F
Feed: TF = 80°F, XF = 10%
Steam: 1000 lb/h, 25 psia
Product: XP = 40%
Assume: zero boiling point rise
cp,solids = 0.35 Btu/lb·°F, cp,w = 1 Btu/lb·°F
Example 8
mV
TV = 120°F

TF = 80°F
XF = 0.1 lb/lb                   TP = 120°F

mF         Juice (120°F)   XP = 0.4 lb/lb
mP = ?

mS
Example 8

   Steam tables:
(hfg)S = 952.16 Btu/lb, at 25 psia (TS = 240°F)
(hg)V = 1113.7 Btu/lb, at 120°F (PV = 1.69 psia)

   Calculate: cp,mix = 0.35· X + 1.0· (1 – X) Btu/lb°F
cpF = 0.935 Btu/lb·°F
cpP = 0.74 Btu/lb·°F
Example 8
mV
TV = 120°F
hg = 1113.7 Btu/lb

TF = 80°F
XF = 0.1 lb/lb                            TP = 120°F
Juice (120°F)     XP = 0.4 lb/lb
mF
cpF = 0.935 Btu/lb°F                         mP = ?
cpF = 0.74 Btu/lb°F
mS
hfg = 952.16 Btu/lb
Example 8

   Solids mass balance:
mF  X F mP  X P
         

   Total mass balance:
mF  mV mP
       

   Total energy balance:
mF  c pF  TF  mS  (h fg ) S  mV  (hg )V  mP  c pP  TP
                                             
Example 8

   Solve for mP:
m S  ( h fg ) S

mP 

c pP T P  R X c pF TF  ( R X 1) ( h g )V

mP = 295 lb/h
Aeration Fan Selection

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)
3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5
DP = (inH2O/ft)design x (depth) + 0.5
4. Total airflow:     cfm = (cfm/bu) x (total bushels)
or:    cfm = (cfm/ ft2) x (floor area)
5. Select fan to deliver flow & pressure (fan data)
Aeration Fan Selection

1.4
1.2
Static Pressure, inH2O

1
0.8                                             System
0.6                                             Fan
0.4
0.2
0
0      500   1000      1500        2000   2500     3000
Airflow, cfm
Aeration Fan Selection
   Example

 Wheat, Kansas, fall aeration
 10,000 bu bin

 16 ft eave height

 pressure aeration system
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)
3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5
4. Total airflow:     cfm = (cfm/bu) x (total bushels)
or:    cfm = (cfm/ ft2) x (floor area)
5. Select fan to deliver flow & pressure (fan data)
Example 9

Recommended Airflow Rates for Dry Grain
(Foster & Tuite, 1982):
Recommended rate*, cfm/bu
Storage             Temperate               Subtropic
Type                 Climate                 Climate
Horizontal         0.05  0.10             0.10  0.20
Vertical           0.03  0.05             0.05  0.10
*Higher rates increase control, flexibility, and cost.
Example 9
Select lowest airflow (cfm/bu) for cooling rate

Approximate Cooling Cycle Fan Time:
Airflow rate (cfm/bu)
Season       0.05       0.10       0.25
Summer      180 hr      90 hr      36 hr
Fall      240 hr     120 hr      48 hr
Winter     300 hr     150 hr      60 hr
Spring     270 hr     135 hr      54 hr
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)

cfm/ft2 = (0.8) x (16 ft) x (0.1 cfm/bu)

cfm/ft2 = 1.3 cfm/ft2
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)

3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5
4. Total airflow:     cfm = (cfm/bu) x (total bushels)
or:    cfm = (cfm/ ft2) x (floor area)
5. Select fan to deliver flow & pressure (fan data)
Pressure drop: DP = (inH2O/ft) x MS x (depth) + 0.5
(note: Ms = 1.3 for wheat)

Airflow Resistance in Grain (Loose-Fill)
100

Soybeans
10                   Corn
2
Airflow, cfm/ft

Barley                          Milo
Wheat
1.3
1

0.1
0.0001     0.001           0.01
0.028   0.1             1   10
Pressure Drop per Foot, inH 2O/ft
Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5

Design Values for Airflow Resistance in Grain
(w/o duct losses)
100

Soybeans
10              Corn
2
Airflow, cfm/ft

Barley                           Milo
Wheat
1.3
1

0.1
0.001           0.01       0.037      0.1                  1   10
Pressure Drop per Foot, inH 2O/ft
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)

3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5

DP = (0.028 inH2O/ft) x 1.3 x (16 ft) + 0.5 inH2O

DP = 1.08 inH2O
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)
3. Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5

DP = (0.037 inH2O/ft) x (16 ft) + 0.5 inH2O

DP = 1.09 inH2O
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)
3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5
4. Total airflow:     cfm = (cfm/bu) x (total bushels)

cfm = (0.1 cfm/bu) x (10,000 bu)
cfm = 1000 cfm
Example 9

1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)
3. Pressure drop: DP = (inH2O/ft)LF x MS x (depth) + 0.5
4. Total airflow:     cfm = (cfm/bu) x (total bushels)
or:    cfm = (cfm/ ft2) x (floor area)
5. Select fan to deliver flow & pressure (fan data)
Example 9

Axial Flow Fan Data (cfm):
Static Pressure, in H2O

Model          0"   0.5"   1"    1.5"   2.5"   3.5"

12"    3/4 hp 1900 1675 1290     815    325    0

12"    1 hp   2308 1963 1460     876    305    0

14"    1.5 hp 3132 2852 2526 2126 1040         0
Example 9

Selected Fan:

12" diameter, ¾ hp, axial flow

Supplies: 1100 cfm @ 1.15 inH2O
(a little extra  0.11 cfm/bu)

Be sure of recommended fan operating range.
Final Thoughts
   Study enough to be confident in your strengths
   Get plenty of rest beforehand
   Calmly attack and solve enough problems to pass