P.E. Review Session
III–D. 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
P.E. Review Session
III. Process Engineering, Part I
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
Knowledge Areas:
I. Common System Applications
II. Natural Resources and Ecology
III. Process Engineering
IV. Facilities
V. Machines
Approx. Exam
Questions
20
15
15
15
15
Current NCEES Topics
Knowledge Areas:
Approx. Exam
Questions
I. Common System Applications
II. Natural Resources and Ecology
20
15
III. Process Engineering
15
IV. Facilities
V. Machines
15
15
Current NCEES Topics
Knowledge Areas:
Approx. Exam
Questions
I. Common System Applications 20
II. Natural Resources and Ecology
15
III. Process Engineering
15
IV. Facilities
V. Machines
15
15
Current NCEES Topics
Primary coverage (Process Engineering):
 I. B.
Energy balances
 III. D.
Mass transfer between phases
 III. I.
Applied psychrometric processes
 III. J.
Mass balances
Also:
 I. P.
Codes, regulations, and standards
 III. E.
Properties of biological materials
Overlaps with (Facilities):
 IV. H, I. Ventilation requirements
Exam
~1%
~1.5%
~1.5%
~1.5%
~1%
~1.5%
~3 %
General area: "Unit Operations"

Within process engineering
Unit 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 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
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
mJ = ? (67% solids)
Illustration – Jam Production
mv = ?
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
m = 0.0025 lb
p
pectin
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
mJ = ? (67% solids)
Illustration – Jam Production
mv = ?
mf = 1 lbfruit (14% solids)
ms = 1.22 lbsugar
mp = 0.0025 lbpectin
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
mJ = ? (67% solids)
Solids Balance:
mf·Csf
Inflow = Outflow + Accumulation
+ 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
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
mJ = ? (67% solids)
Solids Balance:
mf·Csf
Inflow = Outflow + Accumulation
+ 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
Total Mass Balance:
Inflow = Outflow + Accumulation
mf + ms = mv + mJ + 0.0
mJ = ? (67% solids)
Solids Balance:
mf·Csf
Inflow = Outflow + Accumulation
+ 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:
C i
t
Inflow = outflow + accumulation
Chemical
m 1
concentrations:
Ci ,1
2
m
• Energy Balance:
Ci , 2
Energy in = energy out + accumulation
  mass flow rate, kg/s
m
T  temperatur e, K
c p  specific heat capacity, J/kg  K
T
t
m 1
T1
2
m
T2
Principles
• Mass Balance:
Inflow = outflow + accumulation
Chemical
Ci
concentrations:
Ci ,1  m 1  Ci , 2  m 2    V 
t
• Energy Balance:
Energy in = energy out + accumulation
T
m 1  c p  T1  m 2  c p  T2    c p V 
t
(sensible energy)
Principles
• Mass Balance:
Inflow = outflow + accumulation
Chemical
Ci
concentrations:
Ci ,1  m 1  Ci , 2  m 2    V 
t
• Energy Balance:
Energy in = energy out + accumulation
m1·h1
T
m 1  c p  T1  m 2  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
qfrig
Illustration − Apple Cooling
energy in = energy out + accumulation
qin,1+ ... = qout,1+ ... + qa
qfrig
Illustration − Apple Cooling
energy in = energy out + accumulation
qfrig
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/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.)
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,
 = 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 (spores) is reduced 90% (1 log cycle)
 N 
t
   k D  t 
ln 
D
 No 
Sterilization
( 250 FT )
Thermal death time: t  Fo 10

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


z
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
Thermal Death Time Curve
(C. botulinum)
(Esty & Meyer, 1922)
( 250 FT )
t  Fo 10
z
t = thermal death time, min
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
Fo = t @ 250°F (std. temp.)
2.7
Sterilization
10
(Stumbo, 1949, 1953; ...)
N  No ek
D
t
D = decimal reduction time
 N  t
 
ln 
D
 No 
Decimal Reduction Time, min
Thermal Death Rate Plot
1
0.1
0.01
100
110
120
Temperature, °C
130
Sterilization
10
N  No ek
D
t
D = decimal reduction time
 N  t
 
ln 
D
 No 
z
1
Dr = 0.2
(Stumbo, 1949, 1953; ...)
Decimal Reduction Time, min
Thermal Death Rate Plot
0.1
0.01
100
110
121
120
Temperature, °C
130
Sterilization equations
DT  D250 10
D To  T
log

Do
z


 ( 250  T ) 




z
 N o  Fo FT
log 


 N  Do DT
No
Fo  D250 log
N
Fo  t 10


 (T  250 F ) 




z
Fo  t 10


 (T 121C ) 




z
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
DT  D250 10
D To  T
log

Do
z
( 250 F  T )
z
 N o  Fo FT
log 


 N  Do DT
No
Fo  D250 log
N
Fo  t 10
t  Fo 10
(T  250 F )
z
( 250 F  T )
z
Fo  t 10
t  Fo 10
(T 121C )
z
(121C  T )
z
Example 3

If D = 0.25 min at 121°C, find D at 140°C.
z = 10°C.
Example 3
equation
substitute
solve
answer:
log
D To  T

Do
z
D121 = 0.25 min
z = 10°C
D140
121C  140C
log

0.25 min
10C
...
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°C); z = 18°F (10°C)
Example 4
Thermal Death Time Curve
(C. botulinum)
(Esty & Meyer, 1922)
(121 C T )
t  Fo 10
z
t = thermal death time, min
z = DT for 10x change in t, °C
Fo = t @ 121°C (std. temp.)
2.7
Example 4
t  Fo 10
(121 C T )
z
t100  (2.7 min) 10
t100  348 min
(121 C 100 C )
10 C
Heat Exchanger Basics
q  U  Ae  DTm
Heat Exchanger Basics
q  U  Ae  DTm  U  A  DTlm
Heat Exchanger Basics
Dtmax
or
Dtmin
Dtmin
or
Dtmax
q  U  Ae  DTm  U  A  DTlm
DTlm
 DT  DT 
  max DT min  
 ln DTmax

min


 
 (T  T )  (T  T ) 
 (T  T )  (T  T ) 
Hi
Co
Ho
Ci
Ci
Ho
Co 


 Hi

THi  TCo
THi  TCi




ln
ln
T

T
T
Ho
Ci
Ho  TCo

 counter

 parallel


 H cH DTH  m
 C cC DTC  q
m


Heat Exchangers
subscripts:
– hot fluid
C – cold fluid
H
i
o
– side where the fluid enters
– 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
Time Out
Reference Ideas
Need
Mark’s Suggestion

Full handbook


Processing text


Standards


Other text

The one you use regularly
 ASHRAE Fundamentals.
Henderson, Perry, & Young (1997),
Principles of Processing Engineering
 Geankoplis (1993), Transport Processes
& Unit Operations.
ASABE Standards, recent ed.
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: casada@ksu.edu
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
Dtmax
or
Dtmin
Dtmin
or
Dtmax
q  U  Ae  DTm  U  A  DTlm
DTlm
 DT  DT 
  max DT min  
 ln DTmax

min


 
 (T  T )  (T  T ) 
 (T  T )  (T  T ) 
Hi
Co
Ho
Ci
Ci
Ho
Co 


 Hi

THi  TCo
THi  TCi




ln
ln
T

T
T
Ho
Ci
Ho  TCo

 counter

 parallel


 H cH DTH  m
 C cC DTC  q
m


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
90°C

Solution
mf cf DTf = mw cw DTw
60°C
?
20°C
Example 5
90°C

Solution
mf cf DTf = mw cw DTw
60°C
?
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
90°C

Solution
q  U  Ae  DTlm
60°C
 C cC DTC
qm
71°C
20°C
Example 6
DTmin = 90°–60°C
90°C

DTmax = 71°–20°C
Solution
q  U  Ae  DTlm
60°C
 C cC DTC
qm
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
Example 6
DTmin = 90°–60°C
90°C

DTmax = 71°–20°C
Solution
q  U  Ae  DTlm
60°C
 C cC DTC
qm
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·°C
Ae = 1.01 m2
More about Heat Exchangers

Effectiveness ratio (H, P, & Young, pp. 204-212)
(Ta1  Ta 2 )
Ecooling 
,
(Ta1  Tb,in )
UA
NTU 
,
Cmin

One fluid at constant T: R

DTlm correction factors
q  U  A  DTlm  F ( Z , Y )
Cb
R
Ca
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

Latent heat of vaporization

Moisture content: wet and dry basis, and equilibrium
moisture content (ASAE Standard D245.6)

Airflow resistance (ASAE Standard D272.3)
Psychrometrics
Mass Transfer Between Phases
cont.
Equilibrium Moisture
Content, %
25
Effect of temperature on
moisture isotherms (corn data)
20
15
0°C
20°C
40°C
10
5
0
0
20
40
60
Relative Humidity, %
80
100
Mass Transfer Between Phases
Equilibrium Moisture Content, %
cont.
25
20
ASAE Standard D245.6 –
15
Use previous revision (D245.4) for constants
.
or
10
use psychrometric charts in Loewer et al.
5
0°C
20°C
(1994)
40°C
0
0
20
40
60
Relative Humidity, %
80
100
Mass Transfer Between Phases
cont.
Loewer, et al. (1994)
Deep Bed Drying Process
rhe
Twb
rho
TG
To
Equilibrium Moisture Content, %
Use of Moisture Isotherms
Air Temp.
Grain Temp.
Mo
TG
To
Me
rho
Relative Humidity, %
rhe
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)
M wb
1

1  M db
  M db
1

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
; M  dry basis moisture content
MR 
M initial  M equilibrium

Deep bed effects when air flows through more than two
to three layers of kernels.
Drying Process
Constant
Rate
Drying Rate
time varying process
Falling
Rate
Time
Evaporative
Cooling
erh = 100%
aw = 1.0
(Thin-layer)
→
erh < 100%
aw < 1.0
Assume falling rate period, unless…
 Falling rate requires erh or exit air data

Note on water activity
aw
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 activity
aw
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 ≤ 0.85
are sterilized by the controlled aw level.
( not subject to FDA processing regulations;
21 CFR Parts 108, 113, and 114).
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
a
m
humidity ratio : a,in
Fan
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:
 a  t  Da  mg  DWg
m
Basic Drying Process
Mass Conservation

Compare:
moisture added to air
to
moisture removed from product

Total moisture conservation:
kga
kgw
s s kga
kgg
kgw
kgg
 a  t  Da  mg  DWg
m
Basic Drying Process
Mass Conservation – cont’d

Calculate time:

Assumes constant outlet conditions (true initially)


mg DWg
t
 a Da
m
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
cont.
Twb
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%
Texit = 58°F
emc=13%
rhexit
Twb
Texit
Example 7
13%
58
Loewer, et al. (1994)
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 I
Use Loewer, et al. (1994 ) (or ASAE D245.6)

emc=18%
Texit = Tdb,e = TG
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)

emc=18%
Texit = Tdb,e = TG = 53.5°F
Twb
Tdb,e
Cooling Process
Energy Conservation

Compare:
heat added to air
to
heat removed from product

Sensible energy conservation:
 a t ca DTa  mg cg DTg
m
DTg  Tinitial  TII
Cooling Process
Energy Conservation

Compare:
heat added to air
to
heat removed from product

Sensible energy conservation:
 a t ca DTa  mg cg DTg
m

Total energy conservation:
m a t Dha  mg cg DTg
DTg  Tinitial  TII
Cooling Process
(and Drying)
Cooling Process
(and Drying)
erh
Twb
Airflow in Packed Beds
Drying, Cooling, etc.
Design Values for Airflow Resistance in Grain
100
Airflow, cfm/ft2
Soybeans (MS=1.3)
Corn (MS=1.5)
10
Sunflower (MS=1.5)
Milo (MS=1.3)
Wheat (MS=1.3)
1
0.1
0.001
Source: ASABE D272.3, MWPS-29
Barley (MS=1.5)
0.01
0.1
Pressure Drop per Foot, inH2O/ft
1
10
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):
Shedd’s curve multiplier
(Ms = PF = 1.3 to 1.5)
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)
Aeration Fan Selection
Static Pressure, inH2O
1.4
1.2
1
0.8
System
Fan
0.6
0.4
0.2
0
0
500
1000
1500
Airflow, cfm
2000
2500
3000
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)
mV
mF
Juice
mP
mS
Evaporator

Solids mass balance:

Total mass balance:

Total energy balance:
Evaporator

Solids mass balance:
 F  X F m
 P X P
m

X  Concentration, lb
lb
Total mass balance:
m F  m V m P

Total energy balance:
m F  c pF  TF m S  (h fg ) S  m V  hgv m P  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
mF
TP = 120°F
Juice (120°F)
XP = 0.4 lb/lb
mP = ?
mS
Evaporator

Solids mass balance:
 F  X F m
 P X P
m

X  Concentration, lb
lb
Total mass balance:
m F  m V m P

Total energy balance:
m F  c pF  TF m S  (h fg ) S  m V  hgv m P  c pP TP
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
mF
TP = 120°F
Juice (120°F)
XP = 0.4 lb/lb
mP = ?
cpF = 0.935 Btu/lb°F
mS
hfg = 952.16 Btu/lb
cpF = 0.74 Btu/lb°F
Example 8

Solids mass balance:
 F  X F m
 P X P
m

Total mass balance:
 F m
 V m
P
m

Total energy balance:
m F  c pF  TF m S  (h fg ) S  m V  (hg )V m P  c pP TP
Example 8

Solve for mP:
m P 
m S  ( h fg ) S
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:
or:
cfm = (cfm/bu) x (total bushels)
cfm = (cfm/ ft2) x (floor area)
5. Select fan to deliver flow & pressure (fan data)
Aeration Fan Selection
Static Pressure, inH2O
1.4
1.2
1
0.8
System
Fan
0.6
0.4
0.2
0
0
500
1000
1500
Airflow, cfm
2000
2500
3000
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:
or:
cfm = (cfm/bu) x (total bushels)
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
Type
Temperate
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.
Example 9
Select lowest airflow (cfm/bu) for cooling rate
Approximate Cooling Cycle Fan Time:
Season
Summer
Fall
Winter
Spring
Airflow rate (cfm/bu)
0.05
0.10
0.25
180 hr
240 hr
300 hr
270 hr
90 hr
120 hr
150 hr
135 hr
36 hr
48 hr
60 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:
or:
cfm = (cfm/bu) x (total bushels)
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
Airflow, cfm/ft
2
Soybeans
10
Corn
Barley
Milo
Wheat
1.3
1
0.1
0.0001
0.001
0.01
0.028
0.1
Pressure Drop per Foot, inH 2O/ft
1
10
Pressure drop: DP = (inH2O/ft)design x (depth) + 0.5
Design Values for Airflow Resistance in Grain
(w/o duct losses)
100
Airflow, cfm/ft
2
Soybeans
10
Corn
Barley
Milo
Wheat
1.3
1
0.1
0.001
0.01
0.037
0.1
Pressure Drop per Foot, inH 2O/ft
1
10
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:
or:
cfm = (cfm/bu) x (total bushels)
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
M o de l
12"
12"
14"
0"
0.5"
1"
1.5"
2 .5 "
3.5"
81 5
32 5
0
87 6
30 5
0
1.5 hp 3132 2852 2526 2126 1040
0
3/4 hp 1900 1675 1290
1 hp
2308 1963 1460
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