p.e. review session

P.E. Review Session

V–C. Mass Transfer between Phases

Mark Casada, Ph.D., P.E. (M.E.)USDA-ARSCenter for Grain and Animal Health ResearchManhattan, Kansascasada@ksu.edu

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

1,iC2,iC2m

1T2T2mK

J/kg capacity,heat specificc e,temperatur

kg/s rate, flow mass

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.

energy in = energy out + accumulation

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

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

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 = ...

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

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)

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)

)F250(

Sterilization

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

t = thermal death time, min

)F250(

Sterilization

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

)F250(

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

D = decimal reduction time

100 110 120 130

Temperature, °C

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

D = decimal reduction time

100 110 120 130

Temperature, °C

Sterilization equations

250 10

NNDF o

o log250

)1.121(

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

T DD)250(

250 10

NNDF o

o log250

o tF)250(

o tF)1.121(

oFt)250(

oFt)1.121(

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:

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.)

)C1.121(

Example 4

oFt)C1.121(

C10)C100C1.121(

100 10min)7.2(

min 348100 t

Heat Exchanger Basics

me TAUq D

me TAUq D lmTAU D

lmme TAUTAUq DD

T T T T

lnlm T

Hi Co Ho CiT TT T

Hi Ci Ho CoT TT T

max min

( ) ( ) ( ) ( )

counter parallel

qTcmTcm CCCHHH DD

Dtmaxor

Dtminor

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

mf cf DTf = mw cw DTw

Example 5

Solution 90°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

Example 6

Solution

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

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

aacooling C

AUNTUTTTTE

,,)()(

),( 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: 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

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

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)

Effect of temperature onmoisture isotherms (corn data)

0 20 40 60 80 100Relative Humidity, %

0°C20°C40°C

ASAE Standard D245.6 – .

Use previous revision (D245.4) for constants

oruse psychrometric charts in Loewer et al. (1994)

Loewer, et al. (1994)

Effect of temperature onmoisture isotherms (corn data)

0°C20°C40°C

Deep Bed Drying Process

Twb “drying”

Use of Moisture Isotherms

Relative Humidity, %

Air Temp.Grain Temp.

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)

dbwb M

)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

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

grain of mass totalgm

ina, :ratiohumidity

MCgrain in changeD gW

outa, :ratiohumidity

Try it:

Total moisture conservation equation:

ggaa Wmtm DD

Total moisture conservation:

ggaa Wmtm DD

Total moisture conservation:kga

Basic Drying ProcessMass Conservation – cont’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

Time →Co

te FallingRate

erh = 100%

aw = 1.0

erh < 100%

aw < 1.0

EvaporativeCooling

(Thin-layer)

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

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%

Example 7b

Part IUse Loewer, et al. (1994 ) (or ASAE

D245.6)

Texit = Tdb,e = TG = 53.5°FTwb “drying”

emc=18%

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)

Twb “drying”

Airflow in Packed BedsDrying, Cooling, etc.

Source: ASABE D272.3, MWPS-29

0.001 0.01 0.1 1 10

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)

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

mF mP = ?

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

mF mP = ?

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

)()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

SystemFan

Aeration Fan Selection Example

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

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)

0.0001 0.001 0.01 0.1 1 10

/ft2 Corn

Barley Milo

Soybeans

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

Design Values for Airflow Resistance in Grain(w/o duct losses)

0.001 0.01 0.1 1 10

/ft2 Corn

Barley Milo

Soybeans

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

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.

p.e. review session

total mass balance

solids balance

outflow accumulationmf

illustration jam productionjam

mv mj

yield lbjamlbfruit of

heat transfer

flow rate

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