P.E. Review Session

P.E. Review Session

V–C. Mass Transfer between Phases

by

Mark Casada, Ph.D., P.E. (M.E.)USDA-ARSCenter for Grain and Animal Health ResearchManhattan, [email protected]

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 standards1%Overlaps with:

I. D. 2. Applied psychrometric processes~2%

II. A. Environment (Facility Engr.)3-4%

Current NCEES Topics

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; 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)

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 sugarThe mixture is evaporated to 67% soluble solidsWhat 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 + Accumulationmf + ms = mv + mJ + 0.0


mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?



mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?


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 = ?






mJ = ? (67% solids)


ms = 1.22 lbsugar


mv = ?





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

tCi

1m

1,iC2,iC2m

tT

1m

1T2T2mK

KTm

J/kg capacity,heat specificc e,temperatur

kg/s rate, flow mass

p




tCVmCmC i

ii 22,11,

tTVcTcmTcm ppp 2211

(sensible energy)




tCVmCmC i

ii 22,11,

tTVcTcmTcm ppp 2211

(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/dayAmbient design temp: 75°F (loading) decline to 65°F in 20 d…Estimate the refrigeration requirements for the 1st 30 days.

Apple Coolingqfrig

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...


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, Wqr = respiration heat gain from apples,

Wqm = heat from lights, motors, people,

etc., Wqso = solar heat gain through windows,

Wqb = building heat gain through walls,

etc., Wqin = net heat gain from infiltration, Wqe = 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/dayAmbient design temp: 75°F (at loading)

declines to 65°F in 20 daysA = 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.)

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

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)

tko

DeNN

Dttk

NN

D

o

ln

Sterilization

SterilizationThermal 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)

zFtT

o

)F250(

10

Sterilization

Thermal Death Time Curve (C. botulinum)(Esty & Meyer, 1922)

t = thermal death time, min

zFtT

o

)F250(

10

Sterilization


t = thermal death time, minz = DT for 10x change in t, °FFo = t @ 250°F (std. temp.)

z

zFtT

o

)F250(

10

2.7

SterilizationThermal Death Rate Plot(Stumbo, 1949, 1953; ...)

D = decimal reduction time

tko

DeNN

Dt

NN

o

ln

0.01

0.1

1

10

100 110 120 130

Temperature, °C

Dec

imal

Red

uctio

n Ti

me

SterilizationThermal Death Rate Plot(Stumbo, 1949, 1953; ...)

D = decimal reduction time

tko

DeNN

Dt

NN

o

ln

0.01

0.1

1

10

100 110 120 130

Temperature, °C

Dec

imal

Red

uctio

n Ti

me

121.1

Dr =

0.2

z

Sterilization equations

zDDT

T

)250(

250 10

NNDF o

o log250

ztFFT

o

)250(

10

zTT

DD o

o

log

T

T

o

oo

DF

DF

NN

log

ztFCT

o

)1.121(

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

zTF

T DD)250(

250 10

NNDF o

o log250

zFT

o tF)250(

10

zCT

o tF)1.121(

10

zTT

DD o

o

log

T

T

o

oo

DF

DF

NN

log

zTF

oFt)250(

10

zTC

oFt)1.121(

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 oo

log

CCCD

10

1401.121min25.0

log 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 (121.1°C); z = 18°F (10°C)

Example 4


t = thermal death time, minz = DT for 10x change in t, °CFo = t @ 121.1°C (std. temp.)

zFtT

o

)C1.121(

10

2.7

Example 4

zT

oFt)C1.121(

10

C10)C100C1.121(

100 10min)7.2(

t

min 348100 t

Heat Exchanger Basics

me TAUq D


me TAUq D lmTAU D


lmme TAUTAUq DD

DD D

DD

TT T

ln

T T T T

ln

T T T T

lnlm T

T

Hi Co Ho CiT TT T

Hi Ci Ho CoT TT T

Hi Co

Ho Ci

Hi Ci

Ho Co

max min

max

min

( ) ( ) ( ) ( )

counter parallel

qTcmTcm CCCHHH DD

Dtmaxor

Dtmin

Dtminor

Dtmax

Heat Exchangers

subscripts: H – hot fluid i – side where the fluid enters

C – cold fluid o – side where the fluid exitsvariables: 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

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°Cwater, cp = 4.2 kJ/kg°CTfood,inlet = 20°C, Tfood,exit = 60°CTwater,inlet = 90°Cmfood = 0.5 kg/smwater = 1 kg/s

Example 6

Solution

lme TAUq D CCC Tcmq D

90°C

60°C

71°C

20°C

Example 6

Solution


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


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

CRC

AUNTUTTTTE

,,)()(

min,1

21

),( YZFTAUq lm D

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 “Preparing for the Exam”

and/or presentation V-D (Biological Materials)

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

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 100Relative Humidity, %

Equ

ilibr

ium

Moi

stur

e C

onte

nt, %

0°C20°C40°C


cont.

0

5

10

15

20

25


Equ

ilibr

ium

Moi

stur

e C

onte

nt, %

0°C20°C40°C

ASAE Standard D245.6 – .

Use previous revision (D245.4) for constants

oruse psychrometric charts in Loewer et al. (1994)


cont.

Loewer, et al. (1994)


cont.

Effect of temperature onmoisture isotherms (corn data)

0

5

10

15

20

25


Equ

ilibr

ium

Moi

stur

e C

onte

nt, %

0°C20°C40°C

Deep Bed Drying Process

rhe

Twb “drying”

TG To

rho

Use of Moisture Isotherms

Relative Humidity, %

Equ

ilibr

ium

Moi

stur

e 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

11

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

;

Grain Bulk Densityfor deep bed drying calculations

kg/m3 lb/bu[1]

Corn, shelled 721 56Milo (sorghum) 721 56Rice, rough 579 45Soybean 772 60Wheat 772 601Standard bushel. Source: ASAE D241.4

Basic Drying ProcessMass Conservation

Compare: moisture added to airtomoisture removed from product


inaoutaa ,, D

Fan

grain of mass totalgm

ina, :ratiohumidity

MCgrain in changeD gW

outa, :ratiohumidity

am


Try it:

Total moisture conservation equation:


ggaa Wmtm DD


Total moisture conservation:


ggaa Wmtm DD


Total moisture conservation:kga

ss

kgw

kga

kgw

kgg

kgg

Basic Drying ProcessMass Conservation – cont’d

aa

gg

mWm

tD

D

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

Dryi

ng

Rate

Time →Co

nsta

ntRa

te FallingRate

erh = 100%

aw = 1.0

erh < 100%

aw < 1.0

EvaporativeCooling

(Thin-layer)

Drying Processcont.

Twb

Drying Processcont.

Twb “drying”

erhASAE 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 = TG Twb “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 DD

IIinitialg TTT D

Cooling ProcessEnergy Conservation

Compare: heat added to airtoheat removed from product

Sensible energy conservation:

gggaaa TcmTctm DD

Total energy conservation:

gggaa Tcmhtm DDIIinitialg TTT D

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

Airf

low

, cfm

/ft2

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):DP = (inH2O/ft)design x (depth) + 0.5

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

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

mF mP

mV

Juice

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 rise

cp,solids = 0.35 Btu/lb·°F, cp,w = 1 Btu/lb·°F

Example 8

mS

mF mP = ?

mV

Juice (120°F)

TF = 80°F

XF = 0.1 lb/lb TP = 120°F

XP = 0.4 lb/lb

TV = 120°F

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

mF mP = ?

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()(


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)


00.20.40.60.8

11.21.4

0 500 1000 1500 2000 2500 3000Airflow, cfm

Stat

ic Pr

essu

re, i

nH2O

SystemFan





Aeration Fan Selection 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 0.25Summer 180 hr 90 hr 36 hr

Fall 240 hr 120 hr 48 hrWinter 300 hr 150 hr 60 hrSpring 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


Airf

low

, cfm

/ft2 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


Airf

low

, cfm

/ft2 Corn

Barley Milo

Soybeans

Wheat

0.037

1.3

Example 9

DP = (0.028 inH2O/ft) x 1.3 x (16 ft) + 0.5 inH2O

DP = 1.08 inH2O




Example 9

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

DP = 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


2. Airflow: cfm/ft2 = (0.8) x (depth) x (cfm/bu)3. Pressure drop: DP = (inH2O/ft)LF x MS x

(depth) + 0.54. Total airflow: cfm = (cfm/bu) x (total

bushels)

Example 9







Example 9

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111222""" 333///444 hhhppp 111999000000 111666777555 111222999000 888111555 333222555 000

111222""" 111 hhhppp 222333000888 111999666333 111444666000 888777666 333000555 000

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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.





Documents

P.E. Review Session