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									         P.E. Review Session

V–C. Mass Transfer between Phases


   by
   Mark Casada, Ph.D., P.E. (M.E.)
   USDA-ARS
   Center for Grain and Animal Health Research
   Manhattan, Kansas
   casada@ksu.edu
        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).
       Carefully read (and reread) the problem statement.
                   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.
Loading rate: 2000 bu/day
Ambient design temp: 75°F (loading) decline to 65°F in 20 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.
Loading rate: 2000 bu/day
Ambient design temp: 75°F (loading) decline to 65°F in 20 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…
    qs = mload· cpA· ΔT = mload· cpA· ΔT
    qr = mtot· Hresp
    qm = qm1 + qm2 + . . .
    qb = Σ(A/RT)· (Ti – To)
       0
    qin = (Qacpa/vsp)· (Ti – To)
       0
    qso = ...
                   Apple Cooling
   definitions…
     mload = apple loading rate, kg/s (lb/h)
     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.
Loading rate: 2000 bu/day
Ambient design temp: 75°F (at loading)
                               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
qs = mload·cpA·ΔT
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
Additional Example Problems
   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        ...

answer:            D140  0.003 min
                  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
     More about Heat Exchangers

   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
    - emphasize your strengths
    - handle “data look up” problems early
   Plan to figure out some longer or “iffy” problems
    AFTER doing the ones you already know

								
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