Transport phenomena in

food processing

Dr. Sirichai Songsermpong

Dept. of Food Science and Technology

Kasetsart University

Transport phenomena

• Momentum transfer

• Heat transfer

• Mass transfer

Momentum transfer

• From high velocity to low velocity

• Newton’s law

• Shear stress = viscosity*shear rate

• Driving force is the velocity gradient

Heat transfer

• From high temp to low temp

• Fourier’s law of heat conduction

• Heat flux = thermal

conductivity*temperature gradient

• Driving force is the temperature gradient

Mass transfer

• From high concentration to low

concentration

• Fick’s law of mass diffusion

• Mass flux = diffusivity*concentration

gradient

• Driving force is the concentration gradient

Momentum transfer

FLUID

Flow Rate

Velocity Profile

H

W

xzy

SCREW

ND Vz = V cos

Vx = V sin

BARREL

e

Material FlowFlow in pipe

Flow in extruder

Sheeting of dough

Rotational Viscometry

Concentric Cylinders

Sample

Sample

Parallel Plates

M,

M,Ω

m)-(N Torque : M

rad/s Speed,Rotational:Ω

Cone and Plate

Sample

M,

FORCE, FVELOCITY, V

SHEAR FLOW

Moving Plate

y

Stationary PlateArea A

A

Fluid Velocity Profile

A

FStressShear

y

VRateShear

RATESHEAR

STRESSSHEARμVISCOSITY

Viscosity,μ

• Is the flow property of Newtonian fluid

• Resistance of the fluid to flow

• Water has viscosity of 1 cP at 20C

• Milk has viscosity of 2 cP

• 1 cP=10-3 Ns/m2

Effect of Temperature

• Logμ = B/T + C

• T increase, viscosity decrease

Non-Newtonian fluid

• Pseudoplastic (shear thinning)

• Dilatant (shear thickening)

• Bingham plastic

• Casson plastic

Shear Rate

Shear Thickening

Newtonian

Bingham Fluid

Herschel-Bulkley

Rheological Models

Shea

r

Str

ess

Shear Thinning

Power Law Modelnk

o

o

o

n

o k

• Gums/Hydrocolloids

• Emulsions

• Tomato Sauce

• Salad Dressing

• Cream

• Concentrated Protein Solutions

• Starch Suspensions

• Sand + Water

1n

app k

nk

γlog1)(nklogηlog app

γlognkloglog

appηlog

γlog

Slope = n-1

log

γlog

Slope = n

• For both liquids

Pseudoplastic Liquids (n<1) Dilatant Liquids (n>1)

1nk

app kn

Concept of Apparent Viscosity )( app

• Shear Thinning (Pseudoplastic) Liquids n - 1 < 0 (n < 1)

app

app

• Shear Thickening (Dilatant) Liquids n - 1 > 0 (n > 1)

Rheology Measuring Geometries

L2

ΔPRσΔP

Rπ

Q4γQ

3

Capillary Viscometry

Flow Rate

Pressure Difference PΔ

Q

R

L3R

T2T

H

R

Rotational Viscometry

Parallel PlatesωVelocityRotational

Torque T

RH

2

c

2

bc

bc

b

Rh4

RR1TT

RR

R

/

3R2

T3T

Cone and Plates

T and

R

Concentric Cylinders

T and

h

LQPR 8/4

USE OF RHEOLOGICAL DATA

(a) Processing Engineering (Heat Transfer)

Heat Transferred by Natural Convection

Surrounding Fluid

h : Heat Transfer Coefficient

“Newtonian” Fluids

“Non Newtonian” Fluids

31

40140 612751

/

.

BrBZ

.

B

w )PrGL

D(.G.)(

hD

31

43140 00830751

/

/

wrwZ

.

B

w )PrGL

D(.G.)

m

m(

hD

Gz

Gr

Pr

Dimensionless

Numbers

w : Conditions at the wall

B : Bulk Conditions

tyConductiviThermalFluid:

ViscosityFluid:μ

m : Rheological Property

)TT(AhQ surrwall

LD

Fluid flow calculation

• Laminar flow Re<2100

• Turbulent flow Re>4000

• Re=Dvρ/μ

FLUID

Flow Rate

Velocity Profile

Vmax=2vave

Laminar flow

Vmax=1.2Vave

Mass balance for fluid flow

• Continuity equation

• A1v1=A2v2=constant volumetric flow rate

A1

A4

A3

A2

A5

Area

Mass flow rate kg/s

Volumetric flow rate m3/s

Mass/volume=density

Velocity Profile – Laminar Flow

Q

L

rp

2

Velocity Profile u(r)

3

4

4

13

R

Q

n

nw

dr

duw By integration )( rfu

• Rheological Properties have a strong

influence of fluid velocity profile

• Velocity profiles are important in

engineering design, holding tube

calculations, etc.

0

0.0

0.5

1.0

1.5

2.0c=0

u/ume

an

r/R

c = 0.8

c=0.4

Bingham Plastic Velocity Profile

0

0.0

0.5

1.0

1.5

2.0n=1 n=0.8

n=0.4

u/u m

ean

r/R

n=0.1

o

nk

Bingham

Power-Law

4

2

c3

1c

3

41

Rr1c2Rr12

u

u

)/()/(

4

2

c3

1c

3

41

c12

u

u

for r < Ro

for r > Ro

cR

Ro

w

o

Bernouilli’s equation

Energy balance

• Potential energy

• Kinetic energy

• Pressure energy

• Friction loss in pipe

• Mechanical energy from pump

1. Potential energy (J/kg)

ΔPE = g(Z2 - Z1)

2. Kinetic energy

ΔKE =

3. Pressure energy

ΔP/ρ = P2 - P1/ρ

4. Friction energy

Ef = ΔP/ρ

2

2

1

2

2 uu α=0.5 laminar α=1 turbulent

Friction energy

• For straight pipe

• For sudden

contraction

• For sudden

enlargement

D

Luf

PEf

f2

2

2

2uK

Pf

f

f = friction factor(see

graph)

2

2

1

2

1 12

A

AuPf

Kf=0.4(1.25-D22/D1

2)

For fitting (elbows, tees, valves)

• Use equivalent length concept

• Bend and elbow are simply equated to

equivalent length of pipe

• Le=N*D

• Example elbow 90 degree square Le=60D

• See table of friction loss in standard fitting

1

2

L1 L2

L3

L4

L5

Le1

Le1

Le1

Le2

Equivalent Length Concept

Let’s assume that a power-law liquid is flowingnn

Tf

R

Q

n

n

R

kLp

321

4

4

132

2154321 3 LeLeLLLLLLT

Bernouilli’s equation

fEPvgZPvgZ /2//2/ 2

2

221

2

11

fp EP

KEPEE

Power = m(Ep) , m=mass flow rate

Practice

• Apple juice is pumped from an open tank

through 1 in. pipe to a second tank. Mass flow

rate is 1 kg/s through 30 m pipe through 2 90

elbows, 1 angle valve. Compute power

requirement of the pump.

• Given viscosity=2.1*10-3 Pas

• Density=997.1 kg/m3

• Diameter=0.02291 m

• Z1=3 m Z2 = 12 m

• f=0.006

• 90 standard elbow Le=32D

• Angle valve Le=170D

• Solve:Find u=2.433 m/s

• Sudden contraction Kf=0.5

• Friction loss in pipe

• Ep= 9.81(12-3)+2.4332/2+(109.63+1.48)=202.36J/kg

• Power=202.36J/kg*1kg/s=202.36J/s (W)

• With 60%efficiency Power=202.3/.60=337 W

kgJPf

/48.12

)433.2(5.0

2

kgJD

Luf

P/63.1092

2

Heat transfer

• Conduction

• Convection

• Radiation

Conduction

• Fourier’s law of conduction

kAL

TTq

dX

dTkAq

/

21

q=rate of heat transfer (W)

A=cross section area of heat flow (m2)

k=thermal conductivity of the medium (W/mK)

dT/dX= temperature gradient per unit length of path

Practice

• Rectangular slab 1 cm thick

• T1=110C

• T2=90C

• K=17 W/mK

• Heat flux=q/A=?

34,000W/m2

Multilayer sytem

2

2

1

121

k

L

k

L

A

qTT

T1 T2L1

k1

L2

k2

Cylindrical tube

)/ln(

)(2 0

io

ir

rr

TTLkq

Pipe with insulator

• Steam pipe coated

with insulator

2312

31

lmlm

r

kA

r

kA

r

TTq

)/ln(

20

0

i

ilm

rr

rrLA

Convection

• Newton’s law of cooling

ThAq q=rate of heat transfer (W)

h=heat transfer coefficient (W/m2K)

A=heat transfer surface area (m2)

Delta T=difference in temperature between solid surface

and surrounding

Forced convection

• Nu = f(Re, Pr)

• Nu=Nusselt number=hD/k

• Re=Reynolds number=Dvρ/μ

• Pr=Prandtl number=μCp/k

• Many formula in each phenomena

fan Steam

pipeho hi

Free convection

• Nu=a(Gr Pr)m

• Gr=Grashof number=(D3ρ2gβΔT)/μ2

steam

hi

Still air

ho

Overall heat transfer coefficient,

U

• For heat conductance in series

• 1/U=L1/k1+L2/k2+…

L1 L2

Overall heat transfer coefficient

• For convection and conducion

Q Qkha hb

Ta Tb Tc Td

Q=UA(Ta-Td) where

1/U=1/ha+L/k+1/hb

Pipe

• If temperature of fluid is higher, heat flow to

outside

• q=UiAi(Ti-To)

• Ui=overall heat transfer coefficient based on

inside area

olmiiii AhkA

rr

AhAU 0

12 1)(11

Tubular heat exchanger

• q=UiAiΔTlm

)/ln(

)(

12

12

TT

TTTlm

Practice

• Milk (Cp=4 kJ/kgK) flows in inner pipe of heat exchanger. milk enters at 20 C and exits at 60 C. Flow rate 0.5 kg/s.

• Hot water at 90C enters and flow countercurrently at 1 kg/s. Cp of water is 4.18 kJ/kgK.

• Calculate exit temperature of water

• Calculate log mean temperature difference

• If U=2000W/m2K and Di=5 cm calculate L

• Repeat calculation for parallel flow

Answer

• Texit=70.9 C

• Log mean temp difference=39.5 C

• L=6.45m for countercurrent flow

• L=8m for parallel flow

Unsteady state heat transfer

• Temperature changes with time and location

• Important in thermal process

• Governing equation

2

2

x

T

C

k

t

T

p

Biot number

• Bi=Internal resistance to heat transfer/

external resistance to heat transfer

• Bi=(D/k)/(1/h)

• Bi=hD/k

• dimensionless

Biot number

• Bi>40 negligible surface resistance (h

higher than k)

• Bi<0.1 negligible internal resistance (k

higher than h)

• 0.1<Bi<40 finite internal and external

resistance

heat

Negligible internal resistance

(Bi<0.1)

• Heating and cooling of solid metal (high k)

• No temp gradient with location

• Well stirred liquid food in a container

)( TThAdt

dTVCq ap

Ta=Temp of surrounding medium

A=surface area of the object

Negligible internal resistance

(Bi<0.1)

tVChA

ia

a peTT

TT )/(

Practice

• Heating tomato juice from 20 C well stirred

• Surrounding mediumTa=90C

• Kettle radius=0.5 m

• Cp of tomato juice 3.95kJ/kgK

• Density 980 kg/m3

• Time of heating 5 min

• T=? after 5 min of heating (Ans83.3C)

Finite internal and surface

resistance

• 0.1<Bi<40

• Use temp-time chart for sphere, cylinder, slab

• Fourier number Fo

22 D

t

D

t

C

kF

p

o

D is characteristic dimension

Sphere , D is radius

α=thermal diffusivity

Negligible surface resistance

• Bi>40

• Use temp-time chart

• Line k/hD=0

Practice

• Estimate temp at

geometric center of soup

in 303*406 can in boiling

water for 30 min

• Can diameter 0.081m

• Can height 0.11m

• h=2000W/m2K

• Ta=100C

• Ti=35C

• T=30 min=1800s

• Soup properties

• k=0.34 W/mK

• Cp=3.5 kJ/kgK

• ρ=900kg/m3

Answer 48.4C

heat transfer radially

Radiation heat transfer

4TAq

q=rate of heat transfer (W)

E=emissivity (0-1)

σ=Stefan-Boltzmann constant=5.67*10-8 J/sm2K4

A=surface area of object

T=Kelvin temperature

Practice

• How much energy is radiated by this

Infrared source in ten minutes?

• Emissivity=0.8

• Area=5m2

• T=500K

• t=600s

• Answer 8500000 J

Mass transfer

• Evaporation

• Drying

• Distillation

• Evaporative cooler

• Liquid-liquid extraction

• Solid-liquid extraction

• Separation process

• Crystallization

• Gas absorption

Fick’s law

dx

dCDJ

Diffusion flux = (diffusion coefficient) (concentration gradient)

D=diffusion coefficient or diffusivity (m2/s)

Fick’s law

dZ

dXDJ A

J=diffusion flux (mol/m2s)

D=diffusion coefficient or diffusivity (m2/s)

XA=mass fraction of A

Z=position

Osmosis and packaging

• Flux = PA(C2-C1)

C2

C1

P = permeability = Diffusivity*Solubility

A=surface area

C2-C1= differrence in concentration