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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

Current NCEES Topics

- Primary coverage Exam
- V. C. Mass transfer between phases 4
- I. D. 1. Mass and energy balances 2
- Also
- I. B. 1. Codes, regs., and standards 1
- Overlaps with
- I. D. 2. Applied psychrometric processes 2
- II. A. Environment (Facility Engr.) 3-4

Specific Topics/Unit Operations

- Heat mass balance fundamentals
- Evaporation (jam production)
- Postharvest cooling (apple storage)
- Sterilization (food processing)
- Heat exchangers (food cooling)
- Drying (grain)
- Evaporation (juice)
- Postharvest cooling (grain)

Mass Transfer between Phases

- A subcategory of Unit Operations
- Common operations that constitute a process,

e.g. - pumping, cooling, dehydration (drying),

distillation, evaporation, extraction,

filtration, heating, size reduction, and

separation. - How do you decide what unit operations apply to a

particular problem? - Experience is required (practice).
- Carefully read (and reread) the problem statement.

Principles

- Mass Balance
- Inflow outflow accumulation

- Energy Balance
- Energy in energy out accumulation

- Specific equations
- Fluid mechanics, pumping, fans, heat

transfer,drying, separation, etc.

Illustration Jam Production

- Jam is being manufactured from crushed fruit with

14 soluble solids. - Sugar is added at a ratio of 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?

Illustration Jam Production

Illustration Jam Production

Total Mass Balance Inflow Outflow

Accumulation mf ms mv mJ 0.0

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)

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

Illustration Jam Production

What if this was a continuous flow concentrator

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

Principles

- Mass Balance
- Inflow outflow accumulation
- Chemicalconcentrations

- Energy Balance
- Energy in energy out accumulation

Principles

- Mass Balance
- Inflow outflow accumulation
- Chemicalconcentrations

- Energy Balance
- Energy in energy out accumulation

(sensible energy)

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.

Apple Cooling

Principles

- Mass Balance
- Inflow outflow accumulation

- Energy Balance
- Energy in energy out accumulation

- Specific equations
- Fluid mechanics, pumping, fans, heat

transfer,drying, separation, etc.

Illustration - Apple Cooling

energy in energy out accumulation

qin,1 ... qout,1 ... qa

Illustration - Apple Cooling

- Try it...

Illustration - Apple Cooling

- Try it...

An apple orchard produces 30,000 bu of apples a

year, and will store ? 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.

Apple Cooling

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

Apple Cooling

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?

Example 1

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.)

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.)

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?

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

Additional Example Problems

- Sterilization
- Heat exchangers
- Drying
- Evaporation
- Postharvest cooling

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)

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)

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.)

Sterilization

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

Sterilization equations

Sterilization

- Popular problems would be
- Find a new D given change in temperature
- Given one time-temperature sterilization process,

find the new time given another temperature, or

the new temperature given another time

Example 3

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

10C.

Example 3

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

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)

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.)

Example 4

Heat Exchanger Basics

Heat Exchangers

- subscripts H hot fluid i side where the

fluid enters - C cold fluid o side where the fluid exits
- variables m mass flow rate of fluid, kg/s
- c cp heat capacity of fluid,

J/kg-K - C m?c, J/s-K
- U overall heat transfer

coefficient, W/m2-K - A effective surface area, m2
- DTm proper mean temperature

difference, K or C - q heat transfer rate, W
- F(Y,Z) correction factor, dimensionless

Example 5

- A liquid food (cp 4 kJ/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.

Example 5

- Solution

Example 6

- Find the heat exchanger area needed from example

5 if the overall heat transfer coefficient is

2000 W/m2C.

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

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

More about Heat Exchangers

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

- Time Out

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.

Studying for taking the exam

- Practice the kind of problems you plan to work
- Know where to find the data
- See presentation I-C Economics and Statistics, on

Preparing for the Exam

Mass Transfer Between Phases

- Psychrometrics
- A few equations
- Psychrometric charts(SI and English units, high,

low and normal temperatures charts in ASABE

Standards) - Psychrometric Processes Basic Components
- Sensible heating and cooling
- Humidify or de-humidify
- Drying/evaporative cooling

Mass Transfer Between 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)

Mass Transfer Between Phasescont.

Mass Transfer Between Phasescont.

ASAE Standard D245.6

. Use previous revision (D245.4) for

constants or use psychrometric charts in Loewer

et al. (1994)

Mass Transfer Between Phasescont.

Loewer, et al. (1994)

Mass Transfer Between Phasescont.

Deep Bed Drying Process

Use of Moisture Isotherms

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)

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.

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

Basic Drying ProcessMass Conservation

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

Basic Drying ProcessMass Conservation

Fan

Basic Drying ProcessMass Conservation

- Try it
- Total moisture conservation equation

Basic Drying ProcessMass Conservation

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

kgg

s

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)

Drying Processtime varying process

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

Drying Processcont.

Twb

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?

Example 7

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

Example 7

13

Loewer, et al. (1994)

Example 7b

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

Tdb,e

Example 7b

18

53.5

Loewer, et al. (1994)

Example 7b

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

Tdb,e

Cooling ProcessEnergy Conservation

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

- Total energy conservation

Cooling Process(and Drying)

Airflow in Packed BedsDrying, Cooling, etc.

Source ASABE D272.3, MWPS-29

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)

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)

Standards, Codes, Regulations

- Standards
- ASABE
- Already mentioned ASAE D245.6 and D272.3
- ASAE D243.3 Thermal properties of grain and
- ASAE S448 Thin-layer drying of grains and crops
- Several others
- Others not likely for unit operations

- More Examples

Evaporator (Concentrator)

Evaporator

- Solids mass balance
- Total mass balance
- Total energy balance

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

Example 8

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

Example 8

hg 1113.7 Btu/lb

cpF 0.935 Btu/lbF

cpF 0.74 Btu/lbF

hfg 952.16 Btu/lb

Example 8

- Solids mass balance
- Total mass balance
- Total energy balance

Example 8

- Solve for mP
- mP 295 lb/h

Aeration Fan Selection

1. Select lowest airflow (cfm/bu) for cooling

rate 2. Airflow cfm/ft2 (0.8) x (depth) x

(cfm/bu) 3. Pressure drop DP (inH2O/ft)LF x

MS x (depth) 0.5 DP (inH2O/ft)design x

(depth) 0.5 4. Total airflow cfm (cfm/bu) x

(total bushels) or cfm (cfm/ ft2) x (floor

area) 5. Select fan to deliver flow pressure

(fan data)

Aeration Fan Selection

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

Higher rates increase control, flexibility, and

cost.

Example 9 Select lowest airflow (cfm/bu) for

cooling rate

Example 9

1. Select lowest airflow (cfm/bu) for cooling

rate 2. Airflow cfm/ft2 (0.8) x (depth) x

(cfm/bu)

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

cfm/bu) cfm/ft2 1.3 cfm/ft2

Example 9

1. Select lowest airflow (cfm/bu) for cooling

rate 2. Airflow cfm/ft2 (0.8) x (depth) x

(cfm/bu) 3. Pressure drop DP (inH2O/ft)LF x

MS x (depth) 0.5 4. Total airflow cfm

(cfm/bu) x (total bushels) or cfm (cfm/ ft2)

x (floor area) 5. Select fan to deliver flow

pressure (fan data)

Pressure drop DP (inH2O/ft) x MS x (depth)

0.5(note Ms 1.3 for wheat)

Pressure drop DP (inH2O/ft)design x (depth)

0.5

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

Example 9

1. Select lowest airflow (cfm/bu) for cooling

rate 2. Airflow cfm/ft2 (0.8) x (depth) x

(cfm/bu) 3. Pressure drop DP (inH2O/ft)LF x

MS x (depth) 0.5 4. Total airflow cfm

(cfm/bu) x (total bushels)

cfm (0.1 cfm/bu) x (10,000 bu) cfm 1000

cfm

Example 9

1. Select lowest airflow (cfm/bu) for cooling

rate 2. Airflow cfm/ft2 (0.8) x (depth) x

(cfm/bu) 3. Pressure drop DP (inH2O/ft)LF x

MS x (depth) 0.5 4. Total airflow cfm

(cfm/bu) x (total bushels) or cfm (cfm/ ft2)

x (floor area) 5. Select fan to deliver flow

pressure (fan data)

Example 9

Axial Flow Fan Data (cfm)

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.

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