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,

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


I. Common System Applications 20II. Natural Resources and Ecology

15

III. Process Engineering 15IV. Facilities 15V. Machines 15



I. Common System Applications20

II. Natural Resources and Ecology15

III. Process Engineering 15IV. Facilities 15V. Machines 15


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

%


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?


mJ = ? (67% solids)

mf = 1 lbfruit (14% solids)

ms = 1.22 lbsugar

mp = 0.0025 lbpectin

mv = ?


mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?

Total Mass Balance:

Inflow = Outflow + Accumulation

mf + ms = mv + mJ + 0.0


mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?

Total Mass Balance:


mf + ms = mv + mJ + 0.0


mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?

Total Mass Balance:


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 = ? (67% solids)


ms = 1.22 lbsugar


mv = ?

Total Mass Balance:


mf + ms = mv + mJ + 0.0





mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?

Total Mass Balance:


mf + ms = mv + mJ + 0.0




mJ = 2.03 lbJam/lbfruit

mv = 0.19 lbwater/lbfruit


mJ = ? (67% solids)


ms = 1.22 lbsugar


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



t

CVmCmC i

ii 22,11,

t

TVcTcmTcm ppp

2211

(sensible energy)

Principles



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.


qfrig


qfrig

energy in = energy out + accumulation

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


qfrig

energy in = energy out + accumulation

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

Try it -

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


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



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



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


me TAUq lmTAU


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


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)


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


cont.

0

5

10

15

20

25

0 20 40 60 80 100


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)


cont.

Loewer, et al. (1994)

Deep Bed Drying Process

rhe

Twb “drying”

TG To

rho

Use of Moisture Isotherms


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


inaoutaa ,,

Fan

grain of mass totalgm

ina, :ratiohumidity

MCgrain in change gW

outa, :ratiohumidity

am


Try it:

Total moisture conservation equation:


ggaa Wmtm


Total moisture conservation:


ggaa Wmtm


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


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


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:

gggaaa TcmTctm

Total energy conservation:

gggaa Tcmhtm IIinitialg TTT

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)


Pressure drop (loose fill, “Shedd’s data”):

DP = (inH2O/ft)LF x MS x (depth) + 0.5

Pressure drop (design value chart):


0.5 inH2O pressure drop in ducts -Standard design assumption(neglect for full perforated floor)


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


Total mass 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


Total mass 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


Total mass balance:


PPFF XmXm

PVF mmm

PpPPVgVSfgSFpFF TcmhmhmTcm )()(

Example 8

Solve for mP:

mP = 295 lb/h

VgXFpFXPpP

SfgSP hRTcRTc

hmm

)()1(

)(


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)


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


Example

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

Example 9







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



Example 9







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


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


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




Example 9

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

P = 1.09 inH2O



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





Example 9







Example 9

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):

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

Documents

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,