p.e. review session iii–d. mass transfer between phases by mark casada, ph.d., p.e. (m.e.)...
TRANSCRIPT
P.E. Review Session
III–D. Mass Transfer between Phases
by
Mark Casada, Ph.D., P.E. (M.E.)USDA-ARSCenter for Grain and Animal Health ResearchManhattan, [email protected]
P.E. Review Session
III. Process Engineering, Part I
by
Mark Casada, Ph.D., P.E. (M.E.)USDA-ARSCenter for Grain and Animal Health ResearchManhattan, [email protected]
Approx. ExamKnowledge Areas: Questions
I. Common System Applications 20II. Natural Resources and Ecology
15III.Process Engineering 15IV. Facilities 15V. Machines 15
Current NCEES Topics
Approx. ExamKnowledge Areas: Questions
I. Common System Applications 20II. Natural Resources and Ecology
15
III. Process Engineering 15IV. Facilities 15V. Machines 15
Current NCEES Topics
Approx. ExamKnowledge Areas: Questions
I. Common System Applications20
II. Natural Resources and Ecology15
III. Process Engineering 15IV. Facilities 15V. Machines 15
Current NCEES Topics
Primary coverage (Process Engineering):Exam
I. B. Energy balances ~1% III. D. Mass transfer between phases
~1.5% III. I. Applied psychrometric
processes~1.5%
III. J. Mass balances~1.5%
Also: I. P. Codes, regulations, and
standards ~1% III. E. Properties of biological
materials ~1.5%
Overlaps with (Facilities): IV. H, I. Ventilation requirements ~3
%
Current NCEES Topics
General area: "Unit Operations"
Within process engineeringUnit Operations are:
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; these
examples). Carefully read (and reread) the problem
statement.
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)
Processing Textbooks
Henderson, Perry, & Young (1997), Principles of Processing Engineering
Geankoplis (1993), Transport Processes and Unit Operations.
Principles
Mass BalanceInflow = outflow + accumulation
Energy BalanceEnergy in = energy out + accumulation
Specific equationsFluid 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
mJ = ? (67% solids)
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
mv = ?
Illustration – Jam Production
mJ = ? (67% solids)
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
mv = ?
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
Illustration – Jam Production
mJ = ? (67% solids)
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
mv = ?
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
Illustration – Jam Production
mJ = ? (67% solids)
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
mv = ?
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
mJ = ? (67% solids)
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
mv = ?
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
mJ = ? (67% solids)
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
mv = ?
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
mJ = ? (67% solids)
mf = 1 lbfruit (14% 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?
Principles
• Mass Balance:Inflow = outflow + accumulationChemicalconcentrations:
• Energy Balance:Energy in = energy out + accumulation
t
Ci
1m
1,iC2,iC2m
t
T
1m
1T2T2mK
KT
m
J/kg capacity,heat specificc
e,temperatur
kg/s rate, flow mass
p
Principles
• Mass Balance:Inflow = outflow + accumulationChemicalconcentrations:
• Energy Balance:Energy in = energy out + accumulation
t
CVmCmC i
ii 22,11,
t
TVcTcmTcm ppp
2211
(sensible energy)
Principles
• Mass Balance:Inflow = outflow + accumulationChemicalconcentrations:
• Energy Balance:Energy in = energy out + accumulation
t
CVmCmC i
ii 22,11,
t
TVcTcmTcm ppp
2211
(sensible energy) total energy = m·h
m1·h1
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/dayAmbient 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 BalanceInflow = outflow + accumulation
Energy BalanceEnergy in = energy out + accumulation
Specific equationsFluid mechanics, pumping, fans, heat transfer,
drying, separation, etc.
Illustration − Apple Cooling
qfrig
Illustration − Apple Cooling
qfrig
energy in = energy out + accumulation
qin,1+ ... = qout,1+ ... + qa
Illustration − Apple Cooling
qfrig
energy in = energy out + accumulation
qin,1+ ... = qout,1+ ... + qa
Try it -
identify: qin,1 , qin,2 , ...
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/dayAmbient design temp: 75°F (loading) decline to 65°F in 20 d…Estimate the refrigeration requirements for the 1st 30 days.
Apple Cooling
qr
qm
qm
qb
qs
qe
qso
qfrig
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)
qin = (Qacpa/vsp)· (Ti – To)
qso = ...
0
0
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
rA = 46 lb/bu; cpA = 0.9 Btu/lb°F
What is the sensible heat load from the apples on day 3?
Example 1
qr
qm
qm
qb
qs
qe
qso
qfrig
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.)
Note: Ti,avg = 54.5°F
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,r = 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
First order thermal death rate (kinetics) of microbes assumed (exponential decay)
D = decimal reduction time = time, at a given temperature, in
which the number of microbes (spores) is reduced 90% (1 log cycle)
tko
DeNN
D
ttk
N
ND
o
ln
Sterilization
Sterilization
Thermal death time:
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)
zFt
T
o
)F250(
10
Sterilization
Thermal Death Time Curve (C. botulinum)(Esty & Meyer, 1922)
t = thermal death time, min
zFt
T
o
)F250(
10
Sterilization
Thermal Death Time Curve (C. botulinum)(Esty & Meyer, 1922)
t = thermal death time, min
z = DT for 10x change in t, °F
Fo = t @ 250°F (std. temp.)
z
zFt
T
o
)F250(
10
2.7
Sterilization
Thermal Death Rate Plot
(Stumbo, 1949, 1953; ...)
D = decimal reduction time
tko
DeNN
D
t
N
N
o
ln
0.01
0.1
1
10
100 110 120 130
Temperature, °C
Dec
imal
Red
uction
Tim
e, m
in
Sterilization
Thermal Death Rate Plot
(Stumbo, 1949, 1953; ...)
D = decimal reduction time
tko
DeNN
D
t
N
N
o
ln
0.01
0.1
1
10
100 110 120 130
Temperature, °C
Dec
imal
Red
uction
Tim
e, m
in
121
Dr =
0.2
z
Sterilization equations
zDD
T
T
)250(
250 10
N
NDF o
o log250
ztF
FT
o
)250(
10
z
TT
D
D o
o
log
T
T
o
oo
D
F
D
F
N
N
log
ztF
CT
o
)121(
10
Sterilization
Common 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
Sterilization equations
z
TF
T DD)250(
250 10
N
NDF o
o log250
z
FT
o tF)250(
10
z
CT
o tF)121(
10
z
TT
D
D o
o
log
T
T
o
oo
D
F
D
F
N
N
log
z
TF
oFt)250(
10
z
TC
oFt)121(
10
Example 3
If D = 0.25 min at 121°C, find D at 140°C.z = 10°C.
Example 3
equation D121 = 0.25 min
z = 10°C
substitute
solve ...
answer:
zTT
DD o
o
log
C
CCD
10
140121
min25.0log 140
min 003.0140 D
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°C); z = 18°F (10°C)
Example 4
Thermal Death Time Curve (C. botulinum)(Esty & Meyer, 1922)
t = thermal death time, min
z = DT for 10x change in t, °C
Fo = t @ 121°C (std. temp.)
zFt
T
o
)C121(
10
2.7
Example 4
z
T
oFt)C121(
10
C10
)C100C121(
100 10min)7.2(
t
min 348100 t
Heat Exchanger Basics
me TAUq
Heat Exchanger Basics
me TAUq lmTAU
Heat Exchanger Basics
lmme TAUTAUq
TT T
ln
T T T T
ln
T T T T
lnlm T
T
Hi Co Ho Ci
T TT T
Hi Ci Ho Co
T TT T
Hi Co
Ho Ci
Hi Ci
Ho Co
max min
max
min
( ) ( ) ( ) ( )
counter parallel
qTcmTcm CCCHHH
Dtmax
or
Dtmin
Dtmin
or
Dtmax
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 or °C
q = heat transfer rate, W F(Y,Z) = correction factor, dimensionless
Time Out
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 Mark’s Suggestion
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 “PE Exam Study Tips” by Amy
Kaleita Also, “Economics & Statistics”
(Marybeth Lima) under 2011 or 2012 webinars.
Unit ops. questions: [email protected]
Standards, Codes, & Regulations
Standards ASABE
ASAE D245.6 and D272.3 covered in examples
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
Heat Exchangers
lmme TAUTAUq
TT T
ln
T T T T
ln
T T T T
lnlm T
T
Hi Co Ho Ci
T TT T
Hi Ci Ho Co
T TT T
Hi Co
Ho Ci
Hi Ci
Ho Co
max min
max
min
( ) ( ) ( ) ( )
counter parallel
qTcmTcm CCCHHH
Dtmax
or
Dtmin
Dtmin
or
Dtmax
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
Solution
Example 5
Solution
90°C
60°C
?
20°C
mf cf DTf = mw cw DTw
Example 5
Solution
90°C
60°C
?
20°C
mf cf DTf = mw cw DTw
(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
Solution
lme TAUq CCC Tcmq
90°C
60°C
71°C
20°C
Example 6
Solution
lme TAUq CCC Tcmq
90°C
60°C
71°C
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
DTmax = 71°–20°C
DTmin = 90°–60°C
Example 6
Solution
lme TAUq CCC Tcmq
90°C
60°C
71°C
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·°CAe = 1.01 m2
DTmax = 71°–20°C
DTmin = 90°–60°C
More about Heat Exchangers
Effectiveness ratio (H, P, & Young, pp. 204-212)
One fluid at constant T: R DTlm correction factors
a
b
inba
aacooling C
CR
C
AUNTU
TT
TTE
,,)(
)(
min,1
21
),( YZFTAUq lm
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 Sensible heat Latent heat of vaporization
Twb “drying”
Psychrometrics
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.
Effect of temperature onmoisture isotherms (corn data)
0
5
10
15
20
25
0 20 40 60 80 100
Relative Humidity, %
Equ
ilib
rium
Moi
stur
e C
onte
nt, %
0°C20°C40°C
Mass Transfer Between Phases
cont.
0
5
10
15
20
25
0 20 40 60 80 100
Relative Humidity, %
Equ
ilib
rium
Moi
stur
e C
onte
nt, %
0°C20°C40°C
ASAE Standard D245.6 – .
Use previous revision (D245.4) for constants
or
use psychrometric charts in Loewer et al. (1994)
Mass Transfer Between Phases
cont.
Loewer, et al. (1994)
Deep Bed Drying Process
rhe
Twb “drying”
TG To
rho
Use of Moisture Isotherms
Relative Humidity, %
Eq
uil
ibri
um
Moi
stu
re C
onte
nt,
%
Air Temp.Grain Temp.
Me
rho
To
rhe
Mo
TG
DryingDeep 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)
wbdb
dbwb M
MM
M
1
1
1
1
)1()1( 2,21,1 wbwb MWMW
Thin-layer process is not as complex. The common Page eqn. is: (falling rate drying period)
Definitions:k, n = empirical constants (ANSI/ASAE S448.1) t = time
Deep bed effects when air flows through more than two to three layers of kernels.
DryingDeep Bed vs. Thin Layer
ntkeMR ×-=
contentmoisturebasisdryMMM
MMMR
mequilibriuinitial
mequilibriu =-
-= ;
Drying Processtime varying process
Assume falling rate period, unless…
Falling rate requires erh or exit air data
Dry
ing
Rate
Time →C
onst
ant
Rate Falling
Rate
erh = 100%
aw = 1.0
erh < 100%
aw < 1.0
EvaporativeCooling
(Thin-layer)
Note on water activityaw
Definition: aw = erh expressed as a decimal
i.e., 85% erh = 0.85 aw
(recall erh and aw increase with increasing temperature)
Note on water activityaw
Definition: aw = erh expressed as a decimal
i.e., 85% erh = 0.85 aw
(recall erh and aw increase with increasing temperature)
Application:food products with aw ≤ 0.85are sterilized by the
controlled aw level.
( not subject to FDA processing regulations;21 CFR Parts 108, 113, and 114).
Grain Bulk Densityfor 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 ProcessMass Conservation
Compare: moisture added to airtomoisture removed from product
Basic Drying ProcessMass Conservation
inaoutaa ,,
Fan
grain of mass totalgm
ina, :ratiohumidity
MCgrain in change gW
outa, :ratiohumidity
am
Basic Drying ProcessMass Conservation
Try it:
Total moisture conservation equation:
Basic Drying ProcessMass Conservation
ggaa Wmtm
Compare: moisture added to airtomoisture removed from product
Total moisture conservation:
Basic Drying ProcessMass Conservation
ggaa Wmtm
Compare: moisture added to airtomoisture removed from product
Total moisture conservation:kga
ss
kgw
kga
kgw
kgg
kgg
Basic Drying ProcessMass Conservation – cont’d
aa
gg
m
Wmt
Calculate time:
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 Processcont.
Twb
Drying Processcont.
Twb “drying”
erh
ASAE D245.6
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 IIUse Loewer, et al. (1994 ) (or ASAE
D245.6)
RHexit = 55%
Texit = 58°F
Twb“drying”
emc=13%rhexit
Texit
Example 7
Loewer, et al. (1994)
13%
58
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 7b
Part IUse Loewer, et al. (1994 ) (or ASAE
D245.6)
Texit = Tdb,e = TGTwb “drying”
emc=18%
Tdb,e
Example 7b
Loewer, et al. (1994)
18%
53.5
Example 7b
Part IUse Loewer, et al. (1994 ) (or ASAE
D245.6)
Texit = Tdb,e = TG = 53.5°FTwb “drying”
emc=18%
Tdb,e
Cooling ProcessEnergy Conservation
Compare: heat added to airtoheat removed from product
Sensible energy conservation:
gggaaa TcmTctm
IIinitialg TTT
Cooling ProcessEnergy Conservation
Compare: heat added to airtoheat removed from product
Sensible energy conservation:
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Total energy conservation:
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Cooling Process(and Drying)
Cooling Process(and Drying)
Twb“drying”
erh
Airflow in Packed BedsDrying, Cooling, etc.
Source: ASABE D272.3, MWPS-29
0.1
1
10
100
0.001 0.01 0.1 1 10
Air
flo
w, c
fm/f
t2
Pressure Drop per Foot, inH2O/ft
Design Values for Airflow Resistance in Grain
Corn (MS=1.5)
Sunflower (MS=1.5)
Soybeans (MS=1.3)
Barley (MS=1.5)
Wheat (MS=1.3)Milo (MS=1.3)
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
Shedd’s curve multiplier(Ms = PF = 1.3 to 1.5)
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)
Aeration Fan Selection
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 500 1000 1500 2000 2500 3000
Airflow, cfm
Sta
tic
Pre
ssure
, inH
2O
SystemFan
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
More Examples
Evaporator (Concentrator)
mS
mFmP
mV
Juice
Evaporator
Solids mass balance:
Total mass balance:
Total energy balance:
Evaporator
Solids mass balance:
Total mass balance:
Total energy balance:
PPFF XmXm
PVF mmm
PpPPgvVSfgSFpFF TcmhmhmTcm )(
lblbion,ConcentratX
Example 8
Fruit juice concentrator, operating @ T =120°F
Feed: TF = 80°F, XF = 10%
Steam: 1000 lb/h, 25 psiaProduct: XP = 40%
Assume: zero boiling point risecp,solids = 0.35 Btu/lb·°F, cp,w = 1 Btu/lb·°F
Example 8
mS
mFmP = ?
mV
Juice (120°F)
TF = 80°F
XF = 0.1 lb/lb TP = 120°F
XP = 0.4 lb/lb
TV = 120°F
Evaporator
Solids mass balance:
Total mass balance:
Total energy balance:
PPFF XmXm
PVF mmm
PpPPgvVSfgSFpFF TcmhmhmTcm )(
lblbion,ConcentratX
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
mS
mFmP = ?
mV
Juice (120°F)
TF = 80°F
XF = 0.1 lb/lb TP = 120°F
XP = 0.4 lb/lb
TV = 120°F
hg = 1113.7 Btu/lb
hfg = 952.16 Btu/lb
cpF = 0.935 Btu/lb°FcpF = 0.74 Btu/lb°F
Example 8
Solids mass balance:
Total mass balance:
Total energy balance:
PPFF XmXm
PVF mmm
PpPPVgVSfgSFpFF TcmhmhmTcm )()(
Example 8
Solve for mP:
mP = 295 lb/h
VgXFpFXPpP
SfgSP hRTcRTc
hmm
)()1(
)(
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
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 500 1000 1500 2000 2500 3000
Airflow, cfm
Sta
tic
Pre
ssure
, inH
2O
SystemFan
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
StorageType
TemperateClimate
SubtropicClimate
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
cfm/ft2 = (0.8) x (16 ft) x (0.1 cfm/bu)
cfm/ft2 = 1.3 cfm/ft2
1. Select lowest airflow (cfm/bu) for cooling rate
2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)
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)
0.1
1
10
100
0.0001 0.001 0.01 0.1 1 10
Pressure Drop per Foot, inH2O/ft
Air
flo
w, c
fm/f
t2 Corn
Barley Milo
Soybeans
Wheat
0.028
1.3
Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5
Design Values for Airflow Resistance in Grain(w/o duct losses)
0.1
1
10
100
0.001 0.01 0.1 1 10
Pressure Drop per Foot, inH2O/ft
Air
flo
w, c
fm/f
t2 Corn
Barley Milo
Soybeans
Wheat
0.037
1.3
Example 9
P = (0.028 inH2O/ft) x 1.3 x (16 ft) + 0.5 inH2O
P = 1.08 inH2O
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
Example 9
P = (0.037 inH2O/ft) x (16 ft) + 0.5 inH2O
P = 1.09 inH2O
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
Example 9
cfm = (0.1 cfm/bu) x (10,000 bu)
cfm = 1000 cfm
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)
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
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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):
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