Transport phenomena in food processing Dr. Sirichai Songsermpong Dept. of Food Science and Technology Kasetsart University
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