P.E. Review Session

1
P.E. Review Session
VC. 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_at_ksu.edu
2
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

3
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)

4
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.

5
Principles

• Mass Balance
• Inflow outflow accumulation

• Energy Balance
• Energy in energy out accumulation

• Specific equations
• Fluid mechanics, pumping, fans, heat
  transfer,drying, separation, etc.

6
Illustration Jam Production

• Jam is being manufactured from crushed fruit with
  14 soluble solids.
• Sugar is added at a ratio of 5545
• 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?

7
Illustration Jam Production
8
Illustration Jam Production
Total Mass Balance Inflow Outflow
Accumulation mf ms mv mJ 0.0
9
Illustration Jam Production
Total Mass Balance Inflow Outflow
Accumulation mf ms mv mJ 0.0
Solids Balance Inflow Outflow
Accumulation mfCsf msCss mJCsJ
0.0 (1 lb)(0.14lb/lb) (1.22 lb)(1.0lb/lb)
mJ(0.67lb/lb)
10
Illustration Jam Production
Total Mass Balance Inflow Outflow
Accumulation mf ms mv mJ 0.0
Solids Balance Inflow Outflow
Accumulation mfCsf msCss mJCsJ
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
11
Illustration Jam Production
What if this was a continuous flow concentrator
with a flow rate of 10,000 lbfruit/h?
12
Principles

• Mass Balance
• Inflow outflow accumulation
• Chemicalconcentrations

• Energy Balance
• Energy in energy out accumulation

13
Principles

• Mass Balance
• Inflow outflow accumulation
• Chemicalconcentrations

• Energy Balance
• Energy in energy out accumulation

(sensible energy)
14
Illustration - Apple Cooling

• An apple orchard produces 30,000 bu of apples a
  year, and will 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 in 20 d
• Estimate the refrigeration requirements for the
  1st 30 days.

15
Apple Cooling
16
Principles

• Mass Balance
• Inflow outflow accumulation

• Energy Balance
• Energy in energy out accumulation

• Specific equations
• Fluid mechanics, pumping, fans, heat
  transfer,drying, separation, etc.

17
Illustration - Apple Cooling
energy in energy out accumulation
qin,1 ... qout,1 ... qa
18
Illustration - Apple Cooling

• Try it...

19
Illustration - Apple Cooling

• Try it...

An apple orchard produces 30,000 bu of apples a
year, and will 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 in 20 d Estimate the
refrigeration requirements for the 1st 30 days.
20
Apple Cooling
21
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

22
Apple Cooling

• Sensible heat equations
• qs mload cpA ?T mload cpA ?T
• qr mtot Hresp
• qm qm1 qm2 . . .
• qb S(A/RT) (Ti To)
• qin (Qacpa/vsp) (Ti To)
• qso ...

23
Apple Cooling
24
Example 1

• An apple orchard produces 30,000 bu of apples a
  year, and will store ? of the crop in
  refrigerated storage at 31F. Cool to 34F in 5
  day 31F by 10 day.
• Loading rate 2000 bu/day
• Ambient design temp 75F (at loading) declines
  to 65F in 20 days
• rA 46 lb/bu cpA 0.9 Btu/lbF
• What is the sensible heat load from the apples on
  day 3?

25
Example 1
26
Example 1

• qs mloadcpA?T
• mload (2000 bu/day 3 day)(46 lb/bu)
• mload 276,000 lb (on day 3)
• ?T (75F 34F)/(5 day) 8.2F/day
• qs (276,000 lb)(0.9 Btu/lbF)(8.2F/day)
• qs 2,036,880 Btu/day 7.1 ton
• (12,000 Btu/h 1 ton refrig.)

27
Example 1, revisited

• mload 276,000 lb (on day 3)
• Ti,avg (75 74.5 74)/3 74.5F
• ?T (74.5F 34F)/(5 day) 8.1F/day
• qs (276,000 lb)(0.9 Btu/lbF)(8.1F/day)
• qs 2,012,040 Btu/day 7.0 ton
• (12,000 Btu/h 1 ton refrig.)

28
Example 2

• Given the apple storage data of example 1,
• r 46 lb/bu cpA 0.9 Btu/lbF H 3.4
  Btu/lbday
• What is the respiration heat load (sensible) from
  the apples on day 1?

29
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

30
Additional Example Problems

• Sterilization
• Heat exchangers
• Drying
• Evaporation
• Postharvest cooling

31
Sterilization

• 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 is
  reduced 90 (1 log cycle)

32
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 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 4 - 6D
• 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)

33
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 _at_ 250F (std. temp.)

34
Sterilization

• Thermal Death Rate Plot
• (Stumbo, 1949, 1953 ...)
• D decimal reduction time

35
Sterilization equations
36
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

37
Example 3

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

38
Example 3

• equation D121 0.25 min
• z 10C
• substitute
• solve ...
• answer

39
Example 4

• The Fo for a process is 2.7 minutes. What would
  be the processing time if the processing
  temperature was changed to 100C?
• NOTE when only Fo is given, assume standard
  processing conditionsT 250F (121.1C) z
  18F (10C)

40
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 _at_ 121.1C (std. temp.)

41
Example 4
42
Heat Exchanger Basics
43
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

44
Example 5

• A liquid food (cp 4 kJ/kgC) flows in the inner
  pipe of a double-pipe heat exchanger. The food
  enters the heat exchanger at 20C and exits at
  60C. The flow rate of the liquid food is 0.5
  kg/s. In the annular section, hot water at 90C
  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/kgC.

45
Example 5

• Solution

46
Example 6

• Find the heat exchanger area needed from example
  5 if the overall heat transfer coefficient is
  2000 W/m2C.

47
Example 6

• Find the heat exchanger area needed from example
  5 if the overall heat transfer coefficient is
  2000 W/m2C.

Data liquid food, cp 4 kJ/kgC water, cp 4.2
kJ/kgC Tfood,inlet 20C, Tfood,exit
60C Twater,inlet 90C mfood 0.5 kg/s mwater
1 kg/s
48
Example 6
DTmin 9060C

• Solution

DTmax 7120C
q mf cf DTf (0.5 kg/s)(4 kJ/kgC)(60
20C) 80 kJ/s DTlm (DTmax
DTmin)/ln(DTmax/DTmin) 39.6C Ae (80
kJ/s)/(2 kJ/sm2C)(39.5C) 2000 W/m2C 2
kJ/sm2C Ae 1.01 m2
49
More about Heat Exchangers

• Effectiveness ratio (H, P, Young, pp. 204-212)
• One fluid at constant T R??
• DTlm correction factors

50

• Time Out

51
Reference Ideas
Need
Marks 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.

52
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

53
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

54
Mass Transfer Between Phasescont.

• Grain and food drying
• 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)

55
Mass Transfer Between Phasescont.
56
Mass Transfer Between Phasescont.
ASAE Standard D245.6
. Use previous revision (D245.4) for
constants or use psychrometric charts in Loewer
et al. (1994)
57
Mass Transfer Between Phasescont.
Loewer, et al. (1994)
58
Mass Transfer Between Phasescont.
59
Deep Bed Drying Process
60
Use of Moisture Isotherms
61
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)

62
DryingDeep Bed vs. Thin Layer

• Thin-layer process is not as complex. The common
  Page eqn. is (falling rate drying period)
• Definitionsk, n empirical constants
  (ANSI/ASAE S448.1) t time
• Deep bed effects when air flows through more than
  two to three layers of kernels.

63
Grain Bulk Densityfor deep bed drying
calculations
kg/m3 lb/bu1
Corn, shelled 721 56
Milo (sorghum) 721 56
Rice, rough 579 45
Soybean 772 60
Wheat 772 60
1Standard bushel. Source ASAE D241.4 1Standard bushel. Source ASAE D241.4 1Standard bushel. Source ASAE D241.4
64
Basic Drying ProcessMass Conservation

• Compare moisture added to air
• to
• moisture removed from product

65
Basic Drying ProcessMass Conservation
Fan
66
Basic Drying ProcessMass Conservation

• Try it
• Total moisture conservation equation

67
Basic Drying ProcessMass Conservation

• Compare moisture added to air
• to
• moisture removed from product
• Total moisture conservation

kgg
s
68
Basic Drying ProcessMass Conservation contd

• 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)

69
Drying Processtime varying process

• Assume falling rate period, unless
• Falling rate requires erh or exit air data

70
Drying Processcont.
Twb
71
Example 7

• Hard wheat at 75F 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 70F, 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?

72
Example 7

• Part II
• Use Loewer, et al. (1994 ) (or ASAE D245.6)
• RHexit 55
• Texit 58F

73
Example 7
13
Loewer, et al. (1994)
74
Example 7b

• Part I
• Use Loewer, et al. (1994 ) (or ASAE D245.6)
• Texit Tdb,e TG

Tdb,e
75
Example 7b
18
53.5
Loewer, et al. (1994)
76
Example 7b

• Part I
• Use Loewer, et al. (1994 ) (or ASAE D245.6)
• Texit Tdb,e TG 53.5F

Tdb,e
77
Cooling ProcessEnergy Conservation

• Compare heat added to air
• to
• heat removed from product
• Sensible energy conservation

• Total energy conservation

78
Cooling Process(and Drying)
79
Airflow in Packed BedsDrying, Cooling, etc.
Source ASABE D272.3, MWPS-29
80
Aeration Fan Selection

• Pressure drop (loose fill, Shedds data)
• DP (inH2O/ft)LF x MS x (depth) 0.5
• Pressure drop (design value chart)
• DP (inH2O/ft)design x (depth) 0.5

Shedds curve multiplier (Ms PF 1.3 to 1.5)
81
Aeration Fan Selection

• Pressure drop (loose fill, Shedds 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)
82
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

83

• More Examples

84
Evaporator (Concentrator)
85
Evaporator

• Solids mass balance
• Total mass balance
• Total energy balance

86
Example 8

• Fruit juice concentrator, operating _at_ 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

87
Example 8
88
Example 8

• 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 (1 X)
  Btu/lbF
• cpF 0.935 Btu/lbF
• cpP 0.74 Btu/lbF

89
Example 8
hg 1113.7 Btu/lb
cpF 0.935 Btu/lbF
cpF 0.74 Btu/lbF
hfg 952.16 Btu/lb
90
Example 8

• Solids mass balance
• Total mass balance
• Total energy balance

91
Example 8

• Solve for mP
• mP 295 lb/h

92
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)
93
Aeration Fan Selection
94
Aeration Fan Selection

• Example

• Wheat, Kansas, fall aeration
• 10,000 bu bin
• 16 ft eave height
• pressure aeration system

95
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)
96
Example 9
Higher rates increase control, flexibility, and
cost.
97
Example 9 Select lowest airflow (cfm/bu) for
cooling rate
98
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
99
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)
100
Pressure drop DP (inH2O/ft) x MS x (depth)
0.5(note Ms 1.3 for wheat)
101
Pressure drop DP (inH2O/ft)design x (depth)
0.5
102
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
103
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
104
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
105
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)
106
Example 9
Axial Flow Fan Data (cfm)
107
Example 9
Selected Fan 12" diameter, ¾ hp, axial flow
Supplies 1100 cfm _at_ 1.15 inH2O (a little
extra ? 0.11 cfm/bu) Be sure of recommended fan
operating range.
108
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