Mathematical modeling of food processing

939 Pages·2010·28.64 MB·English

by  Mohammed M Farid

1 Fundamentals of Momentum
Transfer
Quang Tuan Pham
University of New South Wales
ContentS
1.1 Basic Concepts: Stress, Strain, and Viscosity ..........................................................................4
1.1.1 Definition of Viscosity ..................................................................................................4
1.1.2 Viscosity Prediction ......................................................................................................5
1.1.2.1 Viscosity Prediction for Gases .......................................................................5
1.1.2.2 Viscosity Prediction for Liquids ....................................................................6
1.1.2.3 Corresponding State Correlation ...................................................................6
1.1.2.4 Viscosity Prediction for Suspensions .............................................................7
1.2 Viscosity Effects in Three-Dimensional Flow .........................................................................7
1.2.1 The Stress Tensor ..........................................................................................................7
1.2.2 The Velocity Gradient Tensor .......................................................................................9
1.2.3 Viscous Stress Correlation in 3-D ..............................................................................11
1.3 The Equations of Change for Fluid Flow ...............................................................................11
1.3.1 The Equation of Continuity ........................................................................................11
1.3.2 The General Momentum Equation .............................................................................12
1.3.3 The Navier Stokes Equation .......................................................................................13
1.3.4 General Transport Equations in Fluids .......................................................................14
1.4 Dimensionless Groups in Fluid Flow .....................................................................................14
1.5 Turbulence Models .................................................................................................................15
1.5.1 Why Turbulence Must be Modeled ............................................................................15
1.5.2 Algebraic Effective Viscosity Models (EVM) ...........................................................17
1.5.3 One-Equation Effective Viscosity Models (EVM) ....................................................17
1.5.4 Two-Equation Effective Viscosity Models (EVM) ....................................................17
1.6 Non-Newtonian Fluids............................................................................................................18
1.6.1 General Observations .................................................................................................18
1.6.2 Extended (Generalized) Newtonian Models ..............................................................19
1.6.2.1 Power Law Models .......................................................................................19
1.6.2.2 Yield Stress Models .....................................................................................20
1.6.3 Time-Dependent Viscosity Models ............................................................................21
1.6.4 Viscoelastic Models ....................................................................................................21
1.6.4.1 Viscoelastic Behavior ...................................................................................21
1.6.4.2 Mechanistic Viscoelastic Models ................................................................22
1.6.4.3 Response to Arbitrary Inputs: the Hereditary Integral ................................25
1.6.4.4 Response to Small Oscillatory Input ...........................................................25
Acknowledgment .............................................................................................................................27
Nomenclature ...................................................................................................................................27
References ........................................................................................................................................29
3
4 Mathematical Modeling of Food Processing
1.1 BaSiC ConCeptS: StreSS, Strain, and ViSCoSity
1.1.1 Definition of Viscosity
The study of fluid dynamics starts with the concept of viscosity. In contrast with solids which tend
to keep their shapes, fluids are free to deform, but not entirely. A deforming fluid is subject to inter-
nal friction forces, arising from the interaction between its molecules as they move relative to each
other.
Imagine two parallel solid plates with a thin layer of fluid of thickness Y sandwiched between
them, the top plate being forced to slide in the x-direction at constant velocity V relative to the
bottom plate (Figure 1.1). Due to molecular attraction, the fluid molecules next to the top plate will
move at the same velocity V as the top plate, while those next to the bottom plate will remain station-
ary (the nonslip condition). A velocity gradient dv/dy will therefore exist in the fluid. In the simplest
x
case, known as Newtonian fluid, this velocity gradient is constant and equal to
dv V
x = (1.1)
dy ∆y
The force necessary to maintain the velocity of the top plate will obviously be proportional to
the area of the plate. It is therefore, more appropriate to talk about the force per unit area, or stress,
τ (measured in Pa or Nm−2):
τ=F/A (1.2)
A larger stress will cause a larger velocity gradient. The viscosity µ of the fluid is defined as the
ratio
τ
µ≡ (1.3)
dv /dy
x
The SI unit for viscosity is Pa.s. The vertical bars indicate that the absolute value must be taken,
i.e., viscosity is always positive. For so called Newtonian fluids, the viscosity is independent of the
velocity gradient (although it may vary according to temperature or pressure).
Equation 1.3 can be written as
d(ρv )
τ=γ x (1.4)
dy
where γ ≡µ/ρ (measured in m2s−1) is called the kinematic viscosity. The stress τ is also called
momentum flux since when you exert a force on an object you give it momentum. The term d(ρv )/dy
x
Moving plate V
Fluid
∆y
Stationary plate
Figure 1.1 Sliding plate experiment.
Fundamentals of Momentum Transfer 5
represents a momentum gradient. The above equation shows that the momentum flux is proportional
to the momentum gradient, just as the heat or mass fluxes are proportional to the temperature or
concentration gradients. Therefore, γ is also known as the diffusivity of momentum (in analogy to
thermal and mass diffusivities). In many situations, the diffusivities of heat, mass and momentum
are of similar magnitude or closely related to each other.
It is stressed that Equation 1.3 holds for the particular case of one-dimensional flow only. Later
we shall generalize this equation to three-dimensional flow.
1.1.2 Viscosity PreDiction
1.1.2.1 Viscosity prediction for gases
The first model for the viscosity of gases is based on the kinetic theory. It assumes that gases are
made up of rigid spherical molecules that move around randomly at an average speed that depends
on temperature. There may be a mean underlying motion in addition to that thermal motion. The
spheres interact with each other by direct contact only, bouncing off each other and any solid surface
in a perfectly elastic manner (no energy loss). The average distance between collisions is called the
mean free path and depends on the spheres’ diameter and number of molecules per unit volume of
space.
If one imagines a fast moving boat passing a stationary boat with passengers jumping back and
forth from one to the other, then it can easily be seen that the passengers jumping from the moving
boat will carry momentum with them and cause the stationary boat to start moving, while the
passengers jumping from the stationary boat will cause the fast moving boat to slow down. This
exchange of momentum can be interpreted as a force between the two boats, in effect a kind of
“viscous friction.” An observer who cannot see the passengers might conclude that there is some
invisible force between the two boats.
Similarly, by considering the exchange of molecules between two layers of gas moving at differ-
ent velocities (Figure 1.2), the resulting momentum exchange (or viscous stress) between these two
layers can be calculated. The rate of molecular exchange depends on the density of the gas and the
thermal velocity or the molecules, which in turn depends on temperature according to the kinetic
theory of gases. Although the molecular movements are random, their effect can be averaged out.
This approach results in the equation
2 κ mT
µ= B (1.5)
3π3/2 d2
where κ is the Boltzmann constant, m the molecular mass, T the temperature and d the diameter.
The above equation shows that the viscosity of gases is independent of pressure. This has been
found to be true for pressures up to about 1 MPa and temperatures above the critical temperature.
However, viscosity is found to increase faster than T. A more accurate prediction equation for gas
viscosity can be obtained from the Chapman–Enskog theory, which discards the assumptions of rigid
V
x2
V
x1
Figure 1.2 Illustration of the kinetic model of viscosity.
6 Mathematical Modeling of Food Processing
n n
o o
ulsi acti
p r
y Re Att
g
r
e
n
E
Distance
Figure 1.3 Lennard–Jones potential for molecular interaction.
spheres and treats the molecular interaction more realistically. This interaction can be expressed as
an intermolecular potential energy field, whose gradient gives the intermolecular force, which may
be repulsive or attractive depending on distance. Using the Lennard–Jones expression for the poten-
tial energy, represented in Figure 1.3, this approach leads to the following equation:
5 κ mT
µ= B (1.6)
16π1/2 σ2Ω
µ
Compared with Equation 1.5, it can be seen that (apart from the constant factor) the rigid sphere
diameter d has been replaced by the so-called collision diameter σ, and Ω is a factor, called the
µ
collision integral for viscosity, which accounts for nonrigid sphere behavior.
1.1.2.2 Viscosity prediction for Liquids
For liquids, the mechanism of viscosity is fundamentally different from that for gases. Here the
molecules are constantly in contact with each other and for a molecule to move relative to the others
it must pass an energy barrier, due to the necessity of squeezing past other molecules. No satisfactory
predictive method is available at the moment, and empirical equations must be used. Also, in contrast
to gases, the viscosity of liquids decrease with temperature according to the empirical equation
B
µ= Aexp  (1.7)
T
where A and B are empirical parameters, since the energy barrier is more easily passed at higher
temperature.
1.1.2.3 Corresponding State Correlation
The behavior of widely different materials can be brought to the same basis by invoking the principle
of corresponding states, according to which reduced properties of gases and liquids follow the same
relationships with reduced pressure and temperature. Reduced means that the value is normalized by
Fundamentals of Momentum Transfer 7
20
Liquid
10
8
6
5 Dense gas
4
c pr = p/pc
µ/µ 3
≡ 25
r 10
µ Two-phase
y 2
sit region 5
o
c
s
d vi 3
e
uc 2
d1.0
Re Critical
0.8 point
1
0.5
0.6
0.5
0.4 pr = 0.2 Low density limit
0.3
0.2
0.4 0.5 0.6 0.8 1.0 2 3 4 5 6 8 10
Reduced temperature T ≡ T/T
r c
Figure 1.4 Plot of reduced viscosity as a function of reduced temperature and pressure. (Reproduced
with permission from Hougen, O.A., Watson, K.M., and Ragatz, R.A., Chemical Process Principles Charts,
Wiley, NY, 1960. Reproduced in Bird, R.B., Stewart, W.E. and Lightfoot, E.N., Transport Phenomena,
Wiley, NY, 2002.)
dividing it by its value at the critical point. The reduced temperature and pressure can be interpreted
as measurements of the deviation from ideal gas behavior, and hence of the interaction between
molecules. Figure 1.4 plots reduced viscosity for many gases and liquids against reduced pressure
and temperature.
1.1.2.4 Viscosity prediction for Suspensions
Many liquid foods are suspensions, for which there is no fundamental way to calculate viscosity.
Various empirical correlations have been proposed and reviewed in Ref. [2].
1.2 ViSCoSity eFFeCtS in three-dimenSionaL FLow
1.2.1 the stress tensor
Consider a very small cube of space in a fluid (Figure 1.5). Each of the six faces will experience a
stress exerted by the surrounding fluid. For example, the face normal to the x-axis will experience a
stress π. This stress is a vector and can be represented by π = (π , π , π )T.
x x xx xy xz
8 Mathematical Modeling of Food Processing
π
y
π
xz
π
y xx
π
x
z
π
xy
x
π
z
Figure 1.5 Stresses on a cube in a fluid.
As the cube shrinks toward a point, Newton’s third law requires that the stresses on opposite
faces of the cube will be equal and opposite. Therefore, the stresses on the cube can be described
by the set of three stress vectors
π = (π , π , π )T
x xx xy xz
π = (π , π , π )T (1.8)
y yx yy yz
π = (π , π , π )T
z zx zy zz
Note: in this chapter, each of the stress vector π denotes the stress experienced by the positive
i
side of the cube (the side with the greater value of x, y or z, in other words the side facing toward +∞).
The stress experienced by its negative side is thus, −π. Some books, such as Bird et al. [2] use the
i
opposite convention. Thus, π > 0 implies a tensile stress on the face normal to x.
xx
The matrix made up of the nine components of the three stress vectors is called the stress
tensor, π:
π π π 
xx yx zx
 
π = π π π (1.9)
 xy yy zy
πxz πyz πzz
In the above tensor, the diagonal elements denote the normal stresses on the faces, while the
non-diagonal elements denote the shear stresses. Each pair of shear stresses on opposite sides causes
a torque. By considering the torque equilibrium on an infinitesimal element of fluid, we can show
that π = π , i.e., the stress tensor is symmetric. Figure 1.6 shows how the stresses on the horizontal
xy yx
faces of the square create a torque, which must be counteracted by an equal an opposite torque due
to the forces on the vertical faces.
The stress tensor is usually written as the sum of a viscous stress tensor τ and a mean normal
stress tensor:
π +π +π
π = τ + xx xx zz I (1.10)
3
Fundamentals of Momentum Transfer 9
F = ∆x∆z π
yx yx
πxy πxy
∆z ∆z
∆y ∆y ∆y
= =
Fxy Fxy
–
∆x
–F = –∆x∆z π
yx yx
Figure 1.6 The torque by shear forces on y-faces,F ∆y = ∆x∆y∆zπ (left) must equal the torque exerted
yx yx
by shear forces on x-faces,F ∆x = ∆x∆y∆zπ (right), henceπ = π .
xy xy yx xy
The pressure p is defined as the negative of the average normal stresses −(π + π = π )/3 (the
xx xx zz
negative sign being due to stresses being defined in such a way that tensile stresses are positive and
compressive stresses are negative), hence the above can also be written
π = τ − pI (1.11)
The viscous stress tensor is also called the deviatoric stress tensor and is, of course, symmetric.
It is to be noted that viscous stress can have a normal component (diagonal terms) as well as shear
components.
1.2.2 the Velocity GraDient tensor
In 3-D, the velocity is a vector v = [v, v, v]T. Let v be the velocity vector at position x = (x, y, z)T
x y z
and v + δv be the velocity vector at a nearby point x + δx = (x + δx, y + δy, z + δz)T. The velocity
gradient is defined as
T T
δv δv δv dv dv dv
lim , ,  =  , ,  (1.12)
δx→0δx δy δz dx dy dz
dv/dx is the rate of change of v in the x-direction, etc. Since v is a vector, it may change in both
magnitude and direction as we move from one point to another. Therefore dv/dx, dv/dy and dv/dz
are all vectors, i.e., the velocity gradient is a vector of vectors, or a tensor. Using the shorthand ∇
(the grad operator) to represent the vector of derivative operators, [∇ = ∂/∂x, ∂/∂y, ∂/∂z]T the velocity
gradient can be written as
∂vx ∂vy ∂vz
 ∂x ∂x ∂x 
 
∇v ≡ ∂vx ∂vy ∂vz (1.13)
 ∂y ∂y ∂y 
 
∂vx ∂vy ∂∂vz
 ∂z ∂z ∂z 
10 Mathematical Modeling of Food Processing
We split the tensor into a symmetric component and an antisymmetric component, then separate
out the mean diagonal term from the symmetric tensor:
 ∂v 1 1∂v ∂v  1∂v ∂v 
 x − e  y + x  z + x
∂x 3 2 ∂x ∂y  2 ∂x ∂zz 
  e 0 00
∇v=12∂∂vxy +∂∂vyx ∂∂vyy −13e 12∂∂vyz ++∂∂vzy+130 e 0
 
0 0 e
1∂v ∂v  1∂v ∂v  ∂v 1   
  z + x  z + y z − e 
2 ∂x ∂z  2 ∂y ∂∂z  ∂z 3 
(1.14)
 1∂v ∂v  1∂v ∂v 
 0  y − x  zz − x
2 ∂x ∂y  2 ∂x ∂z 
 
 1∂v ∂v  1∂v ∂v 
+−  y − x 0  z − y
2 ∂x ∂y  2 ∂yy ∂z 
 
 1∂v ∂v  1∂v ∂vv  
−  z − x  z − y 0 
 2 ∂x ∂z  2 ∂y ∂z  
where e = (∂v/∂x) + (∂v /∂y) + (∂v/∂z) is a term that measures the dilation (expansion) rate. Let us
x y z
give each of the tensors on the right a name: ∇v = D + E + R. Each of these has a physical meaning,
which we state here without proof:
• D measures the deformation rate, i.e., stretching and shear without volume change. It is,
therefore, a generalization of the velocity gradient du/dy in the sliding plate experiment.
We can therefore, expect that it gives rise to viscous stresses.
• E measures the expansion rate. We know that expansion or compression of a fluid gives
rise to changes in pressure, but it may also cause viscous stresses due to the relative motion
of the molecules.
• R (the antisymmetric component) measures the rotation rate. Solid body rotation does
not cause viscous stresses because it does not cause molecules to slide past each other—
therefore it can be dropped from the viscous stress calculation.
Total relative motion in an element of fluid arises from the addition of deformation, expansion
and rotation (see Figure 1.7). Rotation does not cause any viscous stress, which comes mainly from
deformation and (to a much smaller extent) from expansion.
(a) (b)
Expand Deform Rotate
Figure 1.7 Any arbitrary small geometric transformation (a) can be decomposed into a sum of expansion,
(b) deformation and rotation.
Fundamentals of Momentum Transfer 11
1.2.3 Viscous stress correlation in 3-D
From the above discussion we expect the viscous stresses to depend on the deformation tensor D
and expansion tensor E. For a Newtonian fluid, this relationship is linear and we can express the
viscous stress as
τ = aD + bE (1.15)
For the above equation to reduce to Equation 1.3 in the sliding plate experiment, where all terms
except ∂v /∂y are zero, we must have a = 2µ. We also put b/3 = κ where κ is called the bulk viscos-
x
ity. This leads to
 ∂v ∂v ∂v  ∂v ∂v 
 2 x  y + x  z + x
∂x  ∂x ∂y   ∂x ∂z 
  e 0 0
τ =µ∂∂vxyy +∂∂vyx 2∂∂vyy ∂∂vyz +∂∂vzy−23µ−κ0 e 0 (1.16)
 
0 0 e
∂v ∂v  ∂v ∂v  ∂v   
 z + x  z + y 2 z 
 ∂xx ∂z   ∂y ∂z  ∂z 
or
τ =µ∇v+(∇v)T−23µ−κeI (1.17)
For incompressible fluids (e = 0) the second term vanishes:
τ =µ [∇v + (∇v)T] (1.18)
1.3 the equationS oF Change For FLuid FLow
1.3.1 the equation of continuity
The two fundamental transport equations in fluid dynamics are the continuity equation and the momen-
tum equation. The continuity equation expresses conservation of mass in a fluid. It expresses the fact
that the change in mass in a fixed volume of space is equal to the net flows across its surface. By con-
sidering the mass balance over a small cube it may be readily shown that, in Cartesian coordinates
∂ρ ∂(ρv ) ∂(ρv ) ∂(ρv )
=− x − y − z (1.19)
∂t ∂x ∂y ∂z
or in tensor notation
∂ρ (1.20)
=−∇⋅(ρv)
∂t
For incompressible fluids these are reduced to
∂v ∂v ∂v
x + y + z =0 (1.21)
∂x ∂y ∂z
12 Mathematical Modeling of Food Processing
or
∇ ⋅ v = 0 (1.22)
1.3.2 the General MoMentuM equation
The momentum equation expresses conservation of momentum in a fixed volume of space.
Momentum can be transported into or out of the volume in two ways: by convection (momen-
tum carried by fluid entering or leaving the volume) or by molecular forces (pressure and viscous
stresses). Momentum can also be created or lost within the volume because of body forces such as
gravity. By carrying out the momentum balance over a small cube, it can be shown that in Cartesian
coordinates the momentum balance in the i-direction is given by
∂(ρv) ∂τ ∂τ ∂τ  ∂p ∂ρρv v ∂ρv v ∂ρv v 
i = xi + yi + zi − − x i + y i + z i +ρg (1.23)
∂t  ∂x ∂y ∂z  ∂x  ∂x ∂y ∂z  i
i
or in tensor notation
∂(ρv)
=∇⋅τ−∇p−∇⋅(ρvv)+ρg (1.24)
∂t
The symbol τ means the i-th component of the viscous stress on the j-th face. In the above equa-
ji
tions, the left hand side is the rate of change of momentum at a point in space. The terms on the right
hand side express respectively the effects of viscous friction, pressure gradient, momentum convec-
tion, and body forces. It should be noted that the momentum equation is a vector equation, in effect
a set of three equations, one for each coordinate. This is because momentum itself is a vector.
The above form of the momentum equation, where the left hand side describes the rate of momen-
tum change in a fixed volume of space, is called the Eulerian form. An alternative form which may
be more convenient to use in some cases is the Lagrangian form, which expresses the rate of momen-
tum change of a chunk or packet of fluid moving in space, D/Dt. This is also called the substantial
derivative and is related to the local or Eulerian rate of change by
D ∂ ∂ ∂ ∂
= +v +v +v (Cartesian coordinates) (1.25)
Dt ∂t x ∂x y∂y z ∂z
or
D ∂
= +v⋅∇ (tensor notation) (1.26)
Dt ∂t
It can readily be shown that the Lagrangian form of the continuity equation is
Dρ
=−ρ∇⋅v (1.27)
Dt
while a somewhat more complicated proof, involving both the continuity and momentum equations,
leads to the Lagrangian momentum equation:
Dv
ρ =∇⋅τ−∇p+ρg (1.28)
Dt