Download CHAPTER 6

1.3 KINETICS
Mass transport represents a reaction towards the perturbation of a system thermodynamic
equilibrium state. Indeed, as soon as eqs.(4-20) – (4.22) are no longer satisfied, in the attempt
to restore a new equilibrium condition implying a maximum for S (eq.(4.3)) or a minimum
for U (eq.(4.14)), system intensive and extensive parameters are modified and mass transport
is a possible tool to perform this action. In order to have a precise physical idea of what
happens after equilibrium perturbation, let’s focus the attention on Fig. 4.2 showing an
isolated system (in thermodynamic equilibrium) made up by two phases (a, on the left; b, on
the right) constituted by r components and separated by an ideal fixed membrane permeable
to all components. Suppose now to increase the pressure in phase a so that Pa > Pb being
unchanged temperature (Ta = Tb) and components chemical potential ( μ ia  μ ib
i ). Among
the many possible ways the system has to react to this perturbation, the simplest one consists
in pressure re-equilibration with no changes in temperature and chemical potential values
[TProc]. This re-equilibration implies mass motion from phase a to phase b through the
permeable membrane until Pa again equals Pb. Mass flow due to a spatial pressure gradient is
called convection. Analogously, system equilibrium can be broken by modifying (increasing
concentration, for example) the chemical potential μ ia of only one component in phase a
(component i) without altering all other parameters (temperature, pressure and remaining
components chemical potentials). Again, among the possible ways the system has to get a
new equilibrium condition, the simplest one is to render μ ia equal to μ ib without altering all
other parameters. Chemical potential re-equilibration implies the movement of component i
from phase a to phase b crossing the permeable membrane until μ ia = μ ib . In this case,
however, mass transport is the result of the random, thermally induced, Brownian motion of i
molecules as no pressure gradient exists to determine a convective flow. Mass flow induced
by a chemical potential gradient is called diffusion.
1
1.3.1 Mass conservation law
According to the mass conservation law, mass can be neither created nor destroyed so that, in
a fixed space region, mass can increase or decrease only by addition or subtraction from the
surroundings, respectively. Although this law can not be applied in the case of radioisotope
decay or nuclear fission where mass converts into energy according to the well-known
Einstein equation, in the typical pharmaceutical and engineering applications it holds and that
is why it is widely used. In order to transform this law into equations, let’s focus the attention
on a generic flowing fluid composed by r components. In the above mentioned fixed space
region (that, for the sake of simplicity, can be thought as a parallelepiped, see Fig. 4.3), the
conservation law translates into a mass balance that, for the generic ith fluid component can
be expressed as:
rate of i mass  rate of i gain
change
  from surronding

 
 rate of i loss

s  from surronding
 rate of i gain or loss 


s  by reaction

(4.39)
where the last term (called generative term) refers to a generic reaction that can, in principle,
involve component ith and can cause i loss or gain depending on whether i is a reaction
reactant or a product. Obviously, to solve the problem of mass transport, (r - 1) eq.(4.39) like
equations plus an overall mass balance must be written:
rate of total mass  rate of total mass gain
change
  from surronding s

 
 rate of total mass loss 
  from surronding s 
 

(4.41)
where, now, the generative term disappears as the chemical reaction can comport neither a
net mass gain, nor a net mass loss but only a fluid composition change. If X, Y and Z
represent our parallelepiped dimensions in the X, Y and Z Cartesian axes directions,
respectively, the first term of eq.(4.40) will be given by:
rate of i mass  Ci
change
  t X Y Z


(4.42)
2
where t is time, Ci is ith component concentration (mass/volume) in the volume V =
X*Y*Z. The second term in eq.(4.39) can be computed remembering that the incoming
mass flow is given by the sum of mass normally (i.e. in the direction of surface normal)
entering the three surfaces SYZ (area = Y*Z and positioned in X = X; normal = X axis, see
Fig. 4.3), SXZ (area = X*Z and positioned in Y = Y; normal Y axis, see Fig. 4.3) and SXY
(area = X*Y and positioned in Z = Z; normal Z axis, see Fig. 4.3). Accordingly, if JX, JY
and JZ represent the three Cartesian components of the mass flux vector J, we have
rate of total mass gain 
from surronding s
  Y Z J X


X
 X Z J Y
Y
 X Y J Z
(4.43)
Z
where the subscripts X, Y and Z denote that the flux components JX, JY and JZ are evaluated,
respectively, at X, Y and Z. Analogous considerations can be drawn for the third term in
eq.(4.39) with the only difference that, now, we are dealing with an exiting mass flow across
the three surfaces SYZ, SXZ and SXY positioned in (X + X), (Y + Y) and (Z + Z),
respectively (see Fig. 4.3). Accordingly, it follows:
rate of total mass loss 
from surronding s
  Y Z J X


X  ΔX
 X Z J Y
Y  ΔY
 X Y J Z
(4.44)
Z ΔZ
As the explicit expression of the fourth term in eq.(4.39) would require the exact knowledge
of the chemical reaction involving the ith component, for the time being, it will be simply
indicated as Gi*V where Gi indicates the mass of i generated or destroyed per unit time and
volume in virtue of the chemical reaction. Finally, eq.(4.39) becomes:
Ci
V  Y Z J X X  J X
t

X  ΔX
 X Z J
Y Y
 JY
Y  ΔY
 X Y J
Z Z
 JZ
Z ΔZ
 G V
i
(4.45)
If both the left and right hand side terms are divided for V and the limit X, Y and Z  0
is considered, on the basis of partial derivates definition, it follows:
3
Ci
J
J 
 J
  X  Y  Z   Gi
t
Y
Z 
 X
(4.46)
To make eq.(4.46) operative, the mass flux components JX, JY and JZ must be specified.
Remembering that in section 1.3 it was observed that mass transport occurs according to
convection (mass flow is caused by a pressure gradient) and/or diffusion (mass flow is caused
by a chemical potential gradient), it is natural expressing JX, JY and JZ in the following
manner:
J X  Ci vX  Ci Bi
μ i
X
J Y  Ci vY  Ci Bi
μ i
Y
J Z  Ci vZ  Ci Bi
μ i
Z
(4.47)
where vX, vY and vZ are, respectively, the X, Y and Z components of the fluid average
ponderal velocity vector v, while Bi is a coefficient reflecting the mobility of the diffusing
components [Flynn]. While the terms Ci*vX,Y,Z clearly indicate the convection contribute, the
terms containing the partial derivative of the chemical potential represent the contribute of
diffusion (the sign minus indicates that diffusion takes place in the direction of the negative
chemical potential gradient). Assuming ideal solution conditions, the contribute of diffusion
can be replaced by the more usual Fick’s law [TranPhen] according to which eq.(4.47)
becomes:
J X  Ci vX  Dim 
ωi
X
J Y  Ci vY  Dim 
ωi
Y
J Z  Ci vZ  Dim 
ωi
(4.48)
Z
where Dim represents the diffusion coefficient of the ith component in the mixture (our fluid)
while i indicates the ith component mass fraction defined as the ratio between the i mass and
the mixture mass contained in the volume V when X, Y and Z  0. Remembering the
relation between Ci and i:
ωi 
Ci
ρ
(4.49)
where  is fluid density, eq.(4.48) can be expressed explicitly in terms of Ci instead of i:
4
 C C ρ 

J X  Ci vX  Dim  i  i
ρ X 
 X
 C C ρ 

J Y  Ci vY  Dim  i  i
ρ Y 
 Y
 C C ρ 

J Z  Ci vZ  Dim  i  i
ρ Z 
 Z
(4.48’)
Notably, diffusion depends also on density gradient and not only on concentration gradient.
Recalling the definition of the nabla operator vector  (dimensionally a length-1) in Cartesian
coordinates:




i
j
k
X
Y
Z
(4.50)
where i, j and k are, respectively, the X, Y and Z axes versors (unitary vectors defining
direction and versus), eq.(4.46) can be written in a more compact form:
Ci
   J   Gi    Ci v     Dimωi   Gi
t
where Dimω i  Dim
(4.46’)
ω i
ω
ω
i  Dim i j  Dim i k while “  ” indicates the scalar product
X
Y
Z
between vectors. Apart from compactness, the advantage of expressing eq.(4.46) into
eq.(4.46’) form consists in the possibility of using it not only in Cartesian coordinates X, Y, Z,
but also in cylindrical and spherical coordinates once the proper  expression is considered
[transp P]:

1 
R  r  1  t   k
R R
R θ Z
cylinder
(4.51)

1  2
1

sin θ  t  1  p
R r
2
R sin θ  θ
R sin θ  φ
R R
sphere
(4.52)
 
where R is the radial coordinate (versor r),  is the tangential coordinate (versor t) and  is
the angular coordinate (versor p).
In order to write down the mass balance referred to the whole flowing fluid, it is sufficient to
sum up eq.(4.46’) for the r component remembering that:
5
r
C
i 1
i
ρ
(4.53)
ωi  0
(4.54)
0
(4.55)
r
D
im
i 1
r
G
i 1
i
Eq.(4.53) states that fluid density is the sum of each component concentration, eq.(4.54)
affirms that no net fluid motion can take place due to diffusion while eq.(4.55) makes clear
that fluid mass can not modify because of a whatever chemical reaction. Accordingly, the
well-known continuity equation is found:
ρ
   ρv 
t
(4.56)
Notably, this equation can be also derived starting from an overall mass balance similar to
that expressed by eq.(4.46) assuming, Gi = 0, J X  Ci vX , J Y  Ci vY and J Z  Ci vZ .
To conclude this paragraph, it is important to remember that mass diffusion is not only due to
the presence of the concentration gradient (see eqs.(4.46) or (4.48’)), but it can also happen
because of temperature and/or pressure gradients. Indeed, as mass diffusion is due to the
chemical potential gradient which, in turn, is also function of temperature and pressure (see
eqs.(4.8) and (4.37)), it is reasonable that mass diffusion takes place because of concentration
and/or temperature and/or pressure gradient. In addition, other gradients can be responsible
for mass diffusion. Indeed, the presence of an electrical potential gradient provokes ions
diffusion [Transp Proc]. In this case, it is convenient to define the electrochemical potential
~ as follows:
μ
i
~  μ  z Fψ
μ
i
i
i
(4.57)
6
where zi is the net charge on species i, F is Faraday constant and  is the electrostatic
potential. Starting from this definition, after some manipulations, it is possible to derive the
well-known Nerst-Plank equation:
J i   Di Ci  z i Ci
F
ψ
RT
(4.58)
where Ji is the ith component flux, R and T are, respectively, universal gas constant and
absolute temperature. It is clear that the first and second terms on the right represent,
respectively, the fickian and the electrostatic potential gradient contribute to diffusion.
1.3.2 Momentum conservation law
Although r kinetics equations are now available, the problem of getting components profile
concentration in the flowing fluid is not yet solved as the knowledge of the fluid velocity
field is required. Indeed, eq.(4.46’) makes clear (see the term    Ci v ) that the time and
space variation of the fluid velocity vector v must be known to get Ci space and time
dependence  i. To solve this problem, recourse must be made to a particular force balance.
It is well known that the velocity v of a body of mass m (for the sake of simplicity here
thought of spherical shape) changes with time due to the action of all the forces acting on it
according to Newton’s law:
ma  m
dv
  Fi
dt
(4.59)
where a is body acceleration and Fi is the generic force acting on the body. In these terms,
eq.(4.59) holds if m is a constant. When, on the contrary, m is time dependent, eq.(4.59) has
to be generalised. Apart from relativistic effects taking place near light speed, m can be time
dependent also in much more common situations. At this purpose, let’s focus the attention on
a cubic vessel containing an amount M of sand and sliding (see Fig. 4.4) on an inclined plane.
If at time t0 a hole is opened on the vessel back face, sand will progressively go out and M
decreases with time causing a system mass m (sand plus vessel mass) decrease. In order to
7
describe this situation, eq.(4.59) must be generalised asserting that the vector p = mv, called
linear momentum or simply momentum, changes because of the action of all the forces acting
on the vessel and because of sand leaving from the hole. Accordingly, subject of our attention
is no longer body velocity (v) but body momentum (p = mv). This scenario can be further
complicated imagining that vessel motion takes place under a sand rain. Accordingly, if on
one hand, m tends to decrease for sand escaping from the hole, on the other hand, m tends to
increase due to sand coming in through the upper open vessel surface (see Fig.4.4). In
conclusion, the time variation of system momentum (
dp dmv

) depends on the sum of
dt
dt
forces acting on the body plus the incoming and escaping momentum due to in- and out- sand
flow rates, respectively. This simple example can be used to understand what happens to a
fluid element (that, again, can be thought as a parallelepiped of volume V and dimensions
X, Y and Z) of a flowing fluid. Indeed, as in the vessel example, the variation of the fluid
element momentum ( V
ρv
) is due to the action of all the forces acting on it, namely
t
surface and body forces, plus the difference between the incoming and exiting momentum
across parallelepiped surfaces. If the expression of surface and body forces is not so
complicated, as shown later, much more care must be devoted to the expression of the inflow
and outflow momentum. Indeed, as momentum is a vector and it is characterised by three
components, it is convenient, for the time being, to limit our considerations to its X
component per unit volume (the scalar quantity vX can be thought as the momentum X
component concentration per unit volume), being the following argumentations extensible to
the other remaining two components, namely Y (vY) and Z (vZ). The advantage of this
strategy relies on the possibility of dealing with a balance regarding a scalar quantity (vX)
and not a vectorial one (v). Accordingly, as in to the case of the mass conservation law
(paragraph 1.3.1), we are again dealing with the expression of a scalar quantity (vX) flux
8
through a parallelepiped. Consequently, the net balance through the two surfaces (SYZ) of
normal X and placed in X and X+X , reads:
Y Z ρvX vX X  ρvX vX XX 
(4.60)
Similarly, the balances through the two surfaces (SXZ) of normal Y and placed in Y and
Y+Y and the two surfaces (SXY) of normal Z and placed in Z and Z+Z read, respectively:
X Z ρvX vY Y  ρvX vY YY 
(4.61)
X Y ρvX vZ Z  ρvX vZ ZZ 
(4.62)
Thus, the sum of eqs.(4.60)-(4.62) constitutes the net balance of the momentum X component
through parallelepiped surfaces. Now, let’s focus the attention on surface forces acting on the
fluid element and due to the surrounding fluid elements. As the tension (force/surface) state
in a fluid point L is completely described if three stress vectors acting, respectively, on three
mutual perpendicular planes passing through L [Melt] are known, nine scalars (three for each
stress vector) are needed to know the tension state in L. In other words, we need the
knowledge of the stress tensor , whose mathematical expression is:
τ XX
  τ XY
 τ XZ
τ YX
τ YY
τ YZ
τ ZX 
τ ZY 
τ ZZ 
(4.63)
where ij represents the generic tensor element corresponding to the component in the j
direction of the stress vector acting on a plane of normal i. Thus, for example, XX, YX and
ZX represent the X component of the stress vectors acting on a plane whose normal is the X,
Y and Z axis, respectively (see Fig.4.5). It is important to underline that for a resting fluid ij =
0 i, j , this meaning that  is associated exclusively to fluid motion. In addition, it can be
demonstrated [Melt] that  is symmetric, this reflecting in the following relation ij = ji for i
≠ j. In order to be homogeneous with the strategy adopted in writing eqs.(4.60)-(4.62), also in
9
this case only the X component of the three stress vectors will be considered (namely, XX,
YX and ZX). Accordingly, the sum of the X component of the surface forces reads (see
Fig.4.5):

Y Z τ XX
X
 τ XX
XX
  X Z τ
YX Y


 τ YX YY  X Y τ ZX Z  τ ZX
ZZ

(4.64)
Among surface forces, of course, the effect of hydrostatic pressure P has to be considered.
The sum of the X component of surface forces due to P is given by:

Y Z P X  P XX

(4.65)
Again, it is important to underline that this surface force does not depend on fluid motion and
it is present also in a resting fluid. Finally, the most common body force acting on the fluid
element under examination is gravity g, although other body forces due to the presence of
electric and magnetic fields could be similarly considered. The X component of gravity force
reads:
X Y Z ρ g X
(4.66)
Equating the X component of the fluid element momentum variation ( V
ρvX
) to the sum
t
of eqs.(4.60)-(4.62) and eqs.(4.64)-(4.66), dividing both the right and left hand side terms for
V and considering the limit X, Y and Z  0, we get the X component of the generalised
form of eq.(4.59):

ρvX     ρvX vX    ρvX vY    ρvX vZ    τ XX  τ YX  τ ZX
t
Y
Z
Y
Z
 X
  X
 P
 ρg X

 X
(4.67)
Analogously, for the Y and Z components, we have, respectively:

ρvY     ρvY vX    ρvY vY    ρvY vZ    τ XY  τ YY  τ ZY   P  ρg Y
t
Y
Z
Y
Z  Y
 X
  X
(4.68)
10

ρvZ     ρvZ vX    ρvZ vY    ρvZ vZ    τ XZ  τ YZ  τ ZZ
t
Y
Z
Y
Z
 X
  X
 P
 ρg Z

 Z
(4.69)
As for the mass balance, also in this case it is possible to give a more compact expression of
eqs.(4.67)-(4.69):

ρv     ρvv        P  ρg
t
(4.70)
where the left hand side term is the fluid element momentum (per unit volume) time
variation, while the first, second, third and fourth term on the right are, respectively, the net
momentum variation rate due to incoming and exiting mass, the sum of surface forces due to
fluid motion and hydrostatic pressure and the effect of gravitational field. Eq.(4.70) is also
referred to as momentum conservation law as it states that the rate of momentum variation is
due to the sum of forces acting on the fluid element plus the net momentum balance due to
the entering and exiting mass through fluid element surfaces.
The quantity vv is a tensor whose expression is:
vX vX
ρvv  ρ vY vX
 vZ vX
vX vY
vY vY
v Z vY
vX v Z 
vY vZ 
vZ vZ 
Remembering the definition of substantive derivative
(4.71)
D
  :
Dt
D
       v   
Dt
t
(4.72)
and after some manipulations, eq.(4.70) can be recast in the following form:
ρ
D
v        P  ρg
Dt
(4.70’)
Notably, this equation is formally equal to eq.(4.59) as it states that the product of mass (per
unit volume) and acceleration is equal to the sum of all the forces acting on the fluid element.
11
In other words, eq.(4.59) structure can be maintained also for the description of a fluid
element motion provided that traditional acceleration is substituted by a “substantive”
acceleration.
In order to render eq.(4.70) operative, it is necessary to define the relation between the stress
tensor components and the velocity gradient. For a newtonian fluid, in Cartesian coordinates
(see chapter 3), we have:
τ XX  2η
vX  2

  η -    v 
X  3

(4.73)
τ YY  2η
vY  2

  η -    v 
Y  3

(4.74)
τ ZZ  2η
v Z  2

  η -    v 
Z  3

(4.75)
v 
 v
τ XY  τ YX  η X  Y 
 Y X 
(4.76)
v 
 v
τ YZ  τ ZY  η Y  Z 
Y 
 Z
(4.77)
v 
 v
τ ZX  τ XZ  η Z  X 
Z 
 X
(4.78)
where  is fluid shear viscosity, while  is mass viscosity that for gas and liquid is usually
close to zero and, thus, can be neglected [Trans Phen].
Without going in further details, it is sufficient an inspection of eqs.(4.73)-(4.78) to clearly
understand that the stress tensor, and thus velocity, depends on one important fluid property,
namely shear viscosity. In turn, as shear viscosity is temperature dependent (this is true
regardless we are dealing with Newtonian or more complex fluids; see chapter 3), velocity,
stress tensor and thus mass transport depend also on temperature. In addition, also fluid
density depends on temperature. Accordingly, at least theoretically, the r mass balance
12
equations (eqs.(4.46)) and the motion equation (eq.(4.70)) must be coupled to an energy
balance.
1.3.3 Energy conservation law
The fluid occupying the usual element of volume V and dimensions X, Y and Z posses
an amount of energy that can be split into two contributes: the kinetic one, due to fluid
motion, and the internal one due to molecules chaotic Brownian movements, defined U in
paragraph 1.2. Accordingly, the energy balance on V reads:
rate of internal  rate of internal 


 and kinetic
 net rate of
rate of internal  and kinetic

 
 heat received  power dissipated 
and kinetic
  energy gain

  against the
  energy loss


 



 
  due to
energy increase  due to


due
to
surroundin
g
 
 conduction  


convection
 convection

(4.79)
This is nothing more than the first thermodynamic principle for an open system in a variable
regime [TranPhe]. Of course, this balance is incomplete as, for example, some kinds of
energy and mechanism of energy transport, such as nuclear, irradiative and electromagnetic,
have been neglected for the sake of simplicity. Nevertheless, its formulation can be easily
modified to account for these other energy typologies. The rate of internal and kinetic energy
increase (or decrease) can be expressed as:
X Y Z
  ˆ 1 2
 ρU  ρ v 
t 
2

(4.80)
where Û represents the fluid internal energy per unit mass while v is the local fluid velocity
module. While the net internal and kinetic energy income due to convection is expressed by:
 

1
1



Y Z vX  ρUˆ  ρv 2   vX  ρUˆ  ρv 2 

2
2
X

 X  X 
 
 

1
1



 X Z vY  ρUˆ  ρv 2   vY  ρUˆ  ρv 2 

2
2




Y
Y  Y 

13
 

1
1



 X Y vZ  ρUˆ  ρv 2   vZ  ρUˆ  ρv 2 

2
2
Z

 Z Z 
 
(4.81)
the net energy income due to conduction reads:





Y Z qX X  qX XX  X Z qY Y  qY YY  X Y qZ Z  qZ ZZ

(4.82)
where qX, qY and qZ are, respectively, the X, Y and Z components of the conduction energy
flux q whose components can be expressed by the well-known Fourier law:
qX  k
T
X
qY  k
T
Y
qZ  k
T
Z
(4.83)
where k is fluid conductivity. In its motion, the fluid makes work per unit time (power)
against body and surface forces. As, for the sake of simplicity, only gravity has been
considered among the possible body forces, the fluid power dissipated for this reason is:
 ρX Y Z vX g X  vY g Y  vZ g Z 
(4.84)
where the minus remembers that the work per unit time is made against the gravity field
when v and g have opposite versus. The power dissipated due to surface forces regards
pressure P and viscous forces (represented by the stress tensor ) acting on fluid element
faces. For what concerns P, we have:





 Y Z ( PvX ) X  ( PvX ) XX  X Z ( PvY ) Y  ( PvY ) YY  X Y ( PvZ ) Z  ( PvZ ) ZZ

(4.85)
Similarly, for viscous forces, the power dissipated reads:


 Y Z τ XX vX  τ XY vY  τ XZ vZ  X  τ XX vX  τ XY vY  τ XZ vZ  XX 


 X Z τ YXvX  τ YYvY  τ YZvZ  Y  τ YXvX  τ YYvY  τ YZvZ  YY 

 X Y τ ZX vX  τ ZY vY  τ ZZ vZ  Z  τ ZX vX  τ ZY vY  τ ZZ vZ  ZZ

(4.86)
Substituting previous expressions into eq.(4.80) and dividing both members for V in the
limit X, Y and Z  0, the equation energy is found:
14
  ˆ 1 2
 ρU  ρ v  
t 
2

 

 X
  ˆ 1 2      ˆ 1 2      ˆ 1 2  
 vX  ρU  ρv   
 vY  ρU  ρv   
 vZ  ρU  ρv   
2
2
2
  Y  
  Z  
 
 
q
q 
 q
  X  Y  Z   ρvX g X  vY g Y  vZ g Z  
Y
Z 
 X
 PvX PvY PvZ 




Y
Z 
 X
 
τ XX vX  τ XY vY  τ XZ vZ    τ YXvX  τ YYvY  τ YZvZ    τ ZX vX  τ ZY vY  τ ZZ vZ 

Y
Z
 X

(4.87)
Obviously, also this equation can be re-written in a more compact way:

 
  ˆ 1 2
1 2  
 ρU  ρv      ρvUˆ  v      q   ρv  g     Pv       v 
t 
2
2  

 

(4.88)
where the term on the left represents the rate of internal and kinetic energy (per unit volume)
increase (ore decrease), while the right terms are, respectively, the internal and kinetic energy
(per unit volume) income rate due to convection and conduction minus the work (per unit
time and volume) made by the fluid element against gravity, hydrostatic pressure and viscous
forces. Recalling the continuity and momentum equations, after some algebraic
manipulations that are out of the aim of this chapter, eq.(4.88) can be re-written in a more
useful form, i.e explicitly in function of temperature:
ρCˆ p
  ln ρ   DP
DT

   q   
  : v 
Dt
  ln T   P Dt
(4.88’)
 Hˆ 
 ) is fluid specific heat per unit mass, Ĥ represents fluid enthalpy per
where Ĉp (  


T

P
unit mass,    q   k 2T according to Fourier law and  : v  represents fluid heating due
to viscous friction and its expression is:
15
 : v    τ XX vX  τ YY vY  τ ZZ vZ  

X
Y
Z 

v 
v 
v  
 v
 v
 v
  τ XY  X  Y   τ XZ  X  Z   τ YZ  Y  Z  
 Y X 
 Z X 
 Z Y  

(4.89)
In the case of a Newtonian fluid, the stress tensor elements are expressed by eqs.(4.73)-(4.78)
and  : v  becomes:
 : v   η  v
(4.90)
2
2
2
 v  2  v  2  v  2   v
v   v
v   v
v  
 v  2 X    Y    Z     X  Y    X  Z    Y  Z   
 X   Y   Z    Y X   Z X   Z Y  
v
v 
2  v
  X  Y  Z
3  X
Y Z 
2
(4.91)
where  v is the so called dissipation function. Notably, eq.(4.88’) states that the temperature
of a fluid element moving with fluid flow is due to heat conduction (    q ), expansion
  ln ρ   DP

effects (  
) and viscous heating (   : v  ).
  ln T   P Dt
1.3.4 Final considerations
On the bases of what discussed in paragraphs 1.3.1 – 1.3.3, the concentration profile
determination of the r elements constituting our flowing fluid requires to solve r equations
expressing mass conservation law (eq.(4.46’)), three equations expressing momentum
conservation law (eqs.(4.67)-(4.69)) and one equation expressing energy conservation law
(eq.(4.88’). At this purpose, we need some more information regarding fluid properties
dependence on pressure (P), temperature (T) and composition (Ci). In other words, we need
the knowledge of the state equation for fluid density ( = (P, T, Ci)), shear viscosity ( =
(P, T, Ci)), specific heat at constant pressure ( Ĉp = Ĉp (P, T, Ci)), conductivity (k = k (P, T,
Ci) and diffusion coefficients (Di = Di (P, T, Ci)). Once the initial and boundary conditions
16
are defined, the above mentioned equations must be simultaneously solved to get solution.
For the sake of simplicity, in this section, equations have been essentially derived and
discussed referring to Cartesian coordinates system. However, due the vectorial form in
which they have been presented, they can be easily adapted to cylindrical and spherical
coordinates systems (that can be very useful in practical applications) once the expressions of
the various operators are known in each particular coordinates system. The explicit
expressions of these equations in cylindrical and spherical coordinates systems can be found
in specific books [TranPhe].
1.4 FICK EQUATION
Despite the theoretical complexity connected to mass transport, fortunately, in the majority of
pharmaceutical applications, the scenario can be considerably simplified as equations can be
solved separately. In addition, in several cases, while energy balance is not required as
isothermal conditions can be assumed, momentum conservation law becomes trivial as a
uniform and constant velocity field realizes. Indeed, a typical drug delivery system is made
up by three components: a matrix structure (that does not diffuse and, hence, its diffusion
coefficient is zero), water (coming in from the external environment and moving inside
matrix structure) and drug (that, usually, diffuses from the inner matrix to the external release
environment). Accordingly, the problem of drug release can be matched considering only two
equations of the eq.(4.46’) type. Finally, when water absorption is not needed for drug release
or it is very fast in comparison to drug diffusion, only one equation is required. Consequently,
the equation under attention derives from eq.(4.46’) assuming a vanishing velocity field (v =
0) and thus, according to mass conservation law (eq.(4.56), a constant and uniform system
density so that Ci  ωi :
Ci
   DimC i   Gi
t
(4.92)
where Fick law for the matter flux is assumed.
17