Units Operations: Processing and Handling

7/27/2019 Units Operations: Processing and Handling

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P.E. Review Session

III-A. Unit Operations

by

Mark Casada, Ph.D., P.E. (Mech. Engr.)

Lead Scientist and Agricultural EngineerUSDA-ARS

Grain Marketing and Production Research Center

Manhattan, Kansas

[email protected]


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NCEES Topics

Unit Operations 8%

Characterization of Biological Products 4%

Processing & Handling

Mass transfer between phases Principles of unit operations Standards, codes, and regulations Principles of postharvest storage

(Covered previously)


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Unit Operations

What are 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.


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Principles of Unit Operations

Mass BalanceInflow = outflow + accumulation

Energy BalanceEnergy in = energy out + accumulation

Specific equationsFluid mechanics, pumping, fans, heat transfer,

drying, separation, etc.


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IllustrationJam Production

Jam is being manufactured from crushed fruit with14% 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?


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mJ = ? (67% solids)

mf= 1 lbfruit (14% solids)

ms =1.22 lbsugar

mp =0.0025 lbpectin

mv = ?


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mJ = ? (67% solids)


ms =1.22 lbsugar

mp =0.0025 lbpectin

mv = ?

Total Mass Balance:

Inflow = Outflow + Accumulation

mf+ ms = mv + mJ + 0.0


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mJ = ? (67% solids)


ms =1.22 lbsugar

mp =0.0025 lbpectin

mv = ?

Total Mass Balance:

Inflow = Outflow + Accumulationmf+ ms = mv + mJ + 0.0

Solids Balance:


mfCsf + msCss = mJCsJ + 0.0

(1 lb)(0.14lb/lb) + (1.22 lb)(1.0lb/lb) = mJ(0.67

lb/lb)


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mJ = ? (67% solids)


ms =1.22 lbsugar

mp =0.0025 lbpectin

mv = ?

Total Mass Balance:

Inflow = Outflow + Accumulationmf+ ms = mv + mJ + 0.0

Solids Balance:


mfCsf + msCss = mJCsJ + 0.0

(1 lb)(0.14lb/lb) + (1.22 lb)(1.0lb/lb) = mJ(0.67

lb/lb)

mJ = 2.03 lbJam/lbfruit

mw = 0.19 lbwater/lbfruit


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mJ = ? (67% solids)


ms =1.22 lbsugar

mp =0.0025 lbpectin

mv = ?

What if this was a continuous flow concentrator

with a flow rate of 10,000 lbfruit/h?


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Mass Balance:Inflow = outflow + accumulation

Chemical

concentrations:

Energy Balance:

Energy in = energy out + accumulation

t

Ci

1m

1,iC2,iC2m

t

T

1m

1T

2T2mK

KT

m

J/kgcapacity,heatspecificc

e,temperatur

kg/srate,flowmass

p


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Chemical

concentrations:

Energy Balance:


t

C

VmCmCi

ii

22,11,

t

TVcTcmTcm ppp

2211

(sensible energy)


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Chemical

concentrations:

Energy Balance:


t

C

VmCmCi

ii

22,11,

t

TVcTcmTcm ppp

2211

(sensible energy) total energy = mh


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Illustration Apple Cooling

An apple orchard produces 30,000 bu of apples a year, andwill store of the crop in refrigerated storage at 31F.Cool to 34F in 5 d; 31F by 10 d.

Loading rate: 2000 bu/day

Ambient design temp: 75F (loading) decline to 65F in20 d

Estimate the refrigeration requirements for the 1st 30 days.


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Apple Cooling

qfrig


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Mass BalanceInflow = outflow + accumulation

Energy BalanceEnergy in = energy out + accumulation

Specific equationsFluid mechanics, pumping, fans, heat transfer,

drying, separation, etc.


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qfrig


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qfrigenergy in = energy out + accumulation

qin,1+ ... = qout,1+ ... + qa


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qfrigenergy in = energy out + accumulation

qin,1+ ... = qout,1+ ... + qa

Try it -

identify: qin,1 , qin,2 , ...


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Try it...


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Apple Cooling

qr

qm

qm

qb

qsqe

qso

qfrig

qin


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Apple Cooling

Sensible heat terms

qs = sensible heat gain from apples, W

qm = heat from lights, motors, people, etc., Wqso = 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


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Apple Cooling

Sensible heat equations

qs = mload cpA T

qr = mtot Hresp

qm = qm1 + qm2 + . . .

qb = (A/RT) (TiTo)

qin = (Qacpa/vsp) (TiTo)

qso = ...

0

0


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Apple Cooling

definitions

mload = apple loading rate, kg/s (lb/h)

Hresp = sp. rate of heat of respiration, J/kgs (Btu/lbh)mtot = total mass of apples, kg (lb)

cpA = sp. heat capacity of apples, J/kgC (Btu/lbF)cpa = specific heat capacity of air, J/kgC (Btu/lbF)

Qa = volume flow rate of infiltration air, m3/s (cfm)

vsp= specific volume of air, m3/kgDA (ft

3/lbDA)

A = surface area of walls, etc., m2 (ft2)RT = total R-value of walls, etc., m

2C/W (hft2F/Btu)

Ti = air temperature inside,C (F)

To = ambient air temperature,C (F)

qm1

, qm2

= individual mechanical heat loads, W (Btu/h)


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Example 1

An apple orchard produces 30,000 bu of apples a year, and willstore of the crop in refrigerated storage at 31F. Cool to34F in 5 day; 31F by 10 day.

Loading rate: 2000 bu/day

Ambient design temp: 75F (at loading)declines to 65F in 20 day

A = 46 lb/bu; cpA = 0.9 Btu/lb

F

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


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Example 1

qr

qm

qm

qb

qsqe

qso

qfrig

qin


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Example 1

qs = mloadcpAT

mload = (2000 bu/day 3 day)(46 lb/bu)

mload = 276,000 lb (on day 3)

T = (75F 34F)/5 = 8.2F/day

qs = (276,000 lb/day)(0.9 Btu/lbF)(8.2F/day)

qs = 2,036,880 Btu/day = 7.1 ton

(12,000 Btu/h = 1 ton refrig.)


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Example 2

Given the apple storage data of example 1,

= 46 lb/bu; cpA = 0.9 Btu/lbF;

What is the respiration heat load (sensible) from the apples onday 1?


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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/lbday)

qr= 312,800 Btu/day = 1.1 ton


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Unit Operations Examples

Sterilization (food engineering)

Heat exchangers

Evaporation (food engineering) Drying

Postharvest cooling


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First order thermal death rate (kinetics) ofmicrobes assumed (exponential decay)

D = decimal reduction time = time, at a giventemperature, in which the number of microbes is

reduced 90% (1 log cycle)

tko

DeNN

D

ttk

N

ND

o

ln

Sterilization


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Sterilization

The z value is the temperature increase that will result in atenfold increase in death rate The typical z value is 10C (18F) (C. botulinum)

Fo = time in minutes at 250F that will produce the same

degree of sterilization as the given process at temperature T Standard process temp = 250F (121.1C)

Thermal death time: given as a multiple of D Pasteurization: 46D

Milk: 30 min at 62.8C (holder method; old batch method)15 sec at 71.7C (HTST high temp./short time)

Sterilization: 12D

Overkill: 18D (baby food)


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Sterilization

Thermal Death Time Curve(C. botulinum)(Esty & Meyer, 1922)

t= thermal death time, min

z T

oFt)F250(

10


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Sterilization



z = DTfor 10x change in t,

FFo = t@ 250F (std. temp.)

z

z

T

oFt)F250(

10

2.7


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Sterilization

Thermal Death Rate Plot

(Stumbo, 1949, 1953; ...)

D = decimal reduction time

tk

o

D

eNN

D

t

N

N

o

ln

0.01

0.1

1

10

100 110 120 130

Temperature, C

DecimalReductionTime


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Sterilization

Thermal Death Rate Plot

(Stumbo, 1949, 1953; ...)

D = decimal reduction time

tk

o

D

eNN

D

t

N

N

o

ln

0.01

0.1

1

10

100 110 120 130

Temperature, C

DecimalReductionTime

121.1

0.2

z


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Sterilization equations

z

T

T DD)250(

250 10

N

NDF oo log250

z

FT

o tF)250(

10

zCT

o tF)1.121(

10

z

TT

D

D o

o

log

T

T

o

oo

D

F

D

F

N

N

log


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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, orthe new temperature given another time


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Example 3

If D = 0.25 min at 121C, find D at 140C.z = 10C.


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Example 3

equation D121 = 0.25 min

z= 10C

substitute

solve ...

answer:

z

TT

D

D o

o

log

C

CCD

10

1401.121

min25.0log 140

min003.0140 D


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Example 4

The Fo for a process is 2.7 minutes. Whatwould be the processing time if the processing

temperature was changed to 100C?

NOTE: when only Fo is given, assume standard

processing conditions:

T = 250F (121.1C); z = 18F (10C)


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Example 4



z =D

Tfor 10x change in t,

CFo = t@ 121.1C (std. temp.)

zT

oFt)C1.121(

10

2.7


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Example 4

z

T

oFt)C1.121(

10

C10

)C100C1.121(

10min)7.2(

t

min348t


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Heat Exchangers

me TAUq D


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Heat Exchangers

me TAUq D lmTAU D


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Heat Exchangers

lmme TAUTAUq DD

D

D DD

D

TT T

ln

T T T T

ln

T T T T

lnlm T

T

Hi Co Ho Ci

T T

T T

Hi Ci Ho Co

T T

T THi Co

Ho Ci

Hi Ci

Ho Co

max min

max

min

( ) ( ) ( ) ( )

counter parallel

qTcmTcm CCCHHH DD


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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= mc, J/s-K

U= overall heat transfer coefficient, W/m2-K

A = effective surface area, m2

DTm = proper mean temperature difference, K orC

q = heat transfer rate, W

F(Y,Z) = correction factor, dimensionless


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Example 5

A liquid food (cp = 4 kJ/kgC) flows in the inner pipeof a double-pipe heat exchanger. The food enters theheat exchanger at 20C and exits at 60C. The

flow rate of the liquid food is 0.5 kg/s. In the annualsection, hot water at 90C enters the heatexchanger in counter-flow at a flow rate of 1 kg/s.

Assuming steady-state conditions, calculate the exittemperature of the water. The average cp of water is

4.2 kJ/kg

C.


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Example 5

Solution


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Time Out


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Reference Ideas

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.

Need Marks Suggestion

Standards

ASABE Standards, recent ed. Other text Albright (1991),Environment...

Lower et al. (1994), On-Farm Drying and...

MWPS-29 (1999),Dry Grain Aeration

Systems Design Handbook. Ames, IA: MWPS.


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Studying for & taking the exam

Practice the kind of problems you plan towork

Know where to find the data See presentation III-B

(Characterization of Biological Products )


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Example 5

Solution90C

60C

?

20C

wwwfff TcmTcm DD


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Example 6

Find the heat exchanger area needed fromexample 5 if the overall heat transfer coefficient

is 2000 W/m2C.


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Example 6

Find the heat exchanger area needed fromexample 5 if the overall heat transfer coefficient

is 2000 W/m2C.

Data:

liquid food, cp = 4 kJ/kgC

water, cp = 4.18 kJ/kgC

Tfood,inlet = 20

C, Tfood,exit = 60

CTwater,inlet = 90C

mfood = 0.5 kg/s

mwater= 1 kg/s


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Example 6

Solution

lme TAUq D

CCC Tcmq D

90C

60C

71C

20C


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More about Heat Exchangers

Effectiveness ratio (H, P, & Young, pp. 204-212)

One fluid at constant T: R

D

Tlmcorrection factors

a

b

inba

aacooling

C

CR

C

AUNTU

TT

TTE

,,

)(

)(

min,1

21

),( YZFTAUq lm D


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Evaporator (Concentrator)

mS

mFmP

mV

Juice


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Evaporator

Solids mass balance:

Total mass balance:

Total energy balance:

PPFF XmXm

PVF mmm

PpPPgvVSfgSFpFF TcmhmhmTcm )(

lblbion,ConcentratX


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Example 7

Fruit juice concentrator, operating @ T =120F

Feed: TF = 80F, XF = 10%

Steam: 1000 lb/h, 25 psia

Product: XP = 40%

Assume: zero boiling point rise

cp,solids = 0.35 Btu/lbF, cp,w= 1 Btu/lbF


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Example 7

mS

mFmP= ?

mV

Juice(120F)

TF= 80F

XF= 0.1 lb/lbTP= 120F

XP= 0.4 lb/lb

TV= 120F


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Example 7

Steam tables:

(hfg)S= 952.16 Btu/lb, at 25 psia (TS= 240F)

(hg)V= 1113.7 Btu/lb, at 120F (PV= 1.69 psia)

Calculate: cp,mix= 0.35 X + 1.0 (1X)

Btu/lbF

cpF= 0.935 Btu/lbF

cpP= 0.74 Btu/lbF


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Example 7

mS

mFmP= ?

mV

Juice(120F)

TF= 80F

XF= 0.1 lb/lbTP= 120F

XP= 0.4 lb/lb

TV= 120F

hg= 1113.7 Btu/lb

hfg= 952.16 Btu/lb

cpF= 0.935

Btu/lbF cpF= 0.74 Btu/lbF

hlbskgPPFFSSXmXm ,


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Example 7

Solids mass balance:

Total mass balance:

Total energy balance:

PpPPVgVSfgSFpFF TcmhmhmTcm )()(

PPFF XmXm

PVF mmm


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Example 7

Solve formP:

mP

= 295 lb/h

VgXFpFXPpP

SfgSPhRTcRTc

hmm

)()1(

)(


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Mass Transfer Between Phases

Psychrometrics

A few equations

Psychrometric charts(SI and English units, high, low and normal temperatures; chartsinASAE Standards)

Psychrometric ProcessesBasic Components:

Sensible heating and cooling

Humidify or de-humidify

Drying/evaporative cooling

M T f B Ph


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Mass Transfer Between Phasescont.

Grain and food drying

Sensible heat

Latent heat of vaporization

Twb

Psychrometrics

Moisture content: wet and dry basis, and equilibrium

moisture content (ASAE Standard D245.6)Airflow resistance (ASAE Standard D272.3)

M T f B Ph


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0

5

10

15

20

25

0 20 40 60 80 100

Relative Humidity, %

Equilib

riumMoistureContent,%

0C

20C

40C

ASAE Standard D245.6 .Use previous revision (D245.4) for constants

or

use psychrometric charts in Loewer, et al., 1994

M T f B t Ph


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Loewer, et al. (1994)

M T f B t Ph


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Effect of temperature on

moisture isotherms (corn data)

0

5

10

15

20

25

0 20 40 60 80 100


EquilibriumMo

isture

Content,%

0C

20C

40C

D i P


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Drying Processcont.

rhe

Twb

TG To

rho


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Use of Moisture Isotherms


EquilibriumMoistureContent,%

Air Temp.

Grain Temp.

Me

rho

To

rhe

Mo

TG


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Drying (Deep Bed)

Drying grain (e.g., shelled corn) with the drying airflowing through more than two to three layers ofkernels.

Dehydration of solid food materials multiple layers drying & interacting

(single, thin-layer solution is a single equation)

wb

db

db

wbM

MM

M

1

11

1

)1()1( 2,21,1 wbwb MWMW


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Grain Bulk Density

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


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Basic Drying ProcessMass Conservation

Compare: moisture added to air

to

moisture removed from product



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inaoutaa ,, D

Fan

grainofmasstotalgm

ina,:ratiohumidity

MCgraininchangeD gW

outa,:ratiohumidity

am



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Try it:

Total moisture conservation equation:



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ggaa Wmtm DD


to


Total moisture conservation:



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ggaa Wmtm DD


to


Total moisture conservation:kgas

skgwkga

kgwkgg

kgg



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Basic Drying ProcessMass Conservationcontd

aa

gg

m

Wmt

D

D

Calculate time:

Assumes constant outlet conditions

but outlet conditions often change as product dries

use deep-bed drying analysis for non-constant outletconditions(Henderson, Perry, & Young sec. 10.6 for complete analysis)

D i P


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Drying Process

Assume falling rate period, unless

Falling rate requires erh or exit air data

Drying

Rate

Time

Constant

Rate Falling

Rate

erh = 100%aw = 1.0

erh < 100%aw < 1.0

EvaporativeCooling

(Thin-layer)

Drying Process


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Drying Processcont.

Twb

Drying Process


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Drying Processcont.

Twb

erh

ASAE D245.6

E l 8


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Example 8

Hard wheat at 75F is being dried from 18% to12% w.b. in a batch grain drier. Drying will bestopped when the top layer reaches 13%. Ambient

conditions: Tdb = 70F, rh = 20%

Determine the exit air temperature early in thedrying period.

Determine the exit air RH and temperature at theend of the drying period?

E l 8


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Example 8

Part II

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

RHexit = 55%

Texit = 58FTwb

emc=13%rhexit

Texit

E l 8


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Example 8

Loewer, et al. (1994)

13%

E l 8


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Example 8

Part I

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

Texit = Tdb,e = TG = 53.5F

Twb

emc=18%

Tdb,e

Cooling Process


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Cooling ProcessEnergy Conservation

Compare: heat added to air

to

heat removed from product

Sensible energy conservation:

gggaaa TcmTctm DD

IIinitialg TTT D

Cooling Process


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Cooling ProcessEnergy Conservation

Compare: heat added to air

to

heat removed from product

Sensible energy conservation:

gggaaa TcmTctm DD

Total energy conservation:

gggaa Tcmhtm DD

IIinitialg TTT D

Cooling Process


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Cooling Process(and Drying)

Cooling Process


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Cooling Process(and Drying)

Twb

erh

Airflow in Packed Beds


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Airflow in Packed BedsDrying, Cooling, etc.

Source: ASABE D272.3, MWPS-29

Design Values for Airflow Resistance in Grain

0.1

1

10

100

0.001 0.01 0.1 1 10

Pressure Drop per Foot, inH2O/ft

Airflow,cfm/ft

2Corn (PF=1.5)

Sunflower (PF=1.5)

Soybeans (PF=1.3)

Barle PF=1.5

Wheat (PF=1.3)

Milo (PF=1.3)

Standards Codes & Regulations


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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

Fin l Th ht


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Final Thoughts

Study enough to be confident in your strengths

Get plenty of rest beforehand

Calmly attack and solve enough problems to passemphasize your strengths

Plan to figure out some iffy problems AFTER

doing the ones you already know

Aeration Fan Selection


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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) + 0.5DP = (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)



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00.2

0.4

0.6

0.8

1

1.2

1.4

0 500 1000 1500 2000 2500 3000

Airflow, cfm

StaticPressure,inH

2O

System

Fan



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Example

Wheat, Oklahoma, fall aeration

10,000 bu bin 16 ft eave height

pressure aeration system



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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)




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Recommended Air f low Rates for Dry Grain

(Foster & Tuite, 1982):

Recommended rate*, cfm/bu

Storage

TypeTemperate

Climate

Subtropic

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.

Select lowest airflow (cfm/bu) for cooling rate


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Select lowest airflow (cfm/bu) for cooling rate

Approximate Cooling Cycle Fan Time:

Airflow rate (cfm/bu)

Season 0.05 0.10 0.25Summer 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



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

cfm/ft2 = 1.3 cfm/ft2





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



Pressure drop: DP = (inH2O/ft) x MS x (depth) + 0.5(note: M = 1 3 for wheat)


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(note: Ms = 1.3 for wheat)

Airflow Resistance in Grain (Loose-Fill)

0.1

1

10

100

0.0001 0.001 0.01 0.1 1 10


Airflow,cfm/ft2 Corn

Barley Milo

Soybeans

Wheat

0.028

1.3

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


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g

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

0.1

1

10

100

0.001 0.01 0.1 1 10


Airflow,cfm/ft2 Corn

Barley Milo

Soybeans

Wheat

0.037

1.3



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

DP = 1.08 inH2O



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



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

DP = 1.09 inH2O



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



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cfm = (0.1 cfm/bu) x (10,000 bu)

cfm = 1000 cfm




4. Total airflow: cfm = (cfm/bu) x (total bushels)



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SSStttaaatttiiiccc PPPrrreeessssssuuurrreee,,, iiinnn HHH222OOO

MMMooodddeeelll 000""" 000...555""" 111""" 111...555""" 222...555""" 333...555"""

111222""" 333///444 hhhppp 111999000000 111666777555 111222999000 888111555 333222555 000

111222""" 111 hhhppp 222333000888 111999666333 111444666000 888777666 333000555 000

111444""" 111...555 hhhppp 333111333222 222888555222 222555222666 222111222666 111000444000 000

Axial Flow Fan Data (cfm):



110/111


Selected Fan:

12" diameter, hp, axial flow

Supplies: 1100 cfm @ 1.2 inH2O

(a little extra 0.11 cfm/bu)

Be sure of recommended operating range.


111/111

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Units Operations: Processing and Handling