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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