3.1 Bilancio Di Massa
Transcript of 3.1 Bilancio Di Massa
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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 lon!er satisfied, in the attempt
to restore a new equilibrium condition implyin! a ma"imum for S(eq.(4.#)) or a minimum
for U(eq.(4.$4)), system intensi%e and e"tensi%e parameters are modified and mass transport
is a possible tool to perform this action. In order to ha%e a precise physical idea of what
happens after equilibrium perturbation, let&s focus the attention on 'i!. 4.2 showin! an
isolated system(in thermodynamic equilibrium) made up by two phases (a, on the left b, on
the ri!ht) constituted by rcomponents and separated by an ideal fi"ed membrane permeable
to all components. uppose now to increase the pressure in phase a so thatPa * Pbbein!
unchan!ed temperature (Ta+ Tb) and components chemical potential ( i,, b
i
a
i = ). mon!
the many possible ways the system has to react to this perturbation, the simplest one consists
in pressure re-equilibration with no chan!es in temperature and chemical potential %alues
/roc1. /his re-equilibration implies mass motion from phase a to phase b throu!h the
permeable membrane untilPaa!ain equalsPb. Mass flow due to a spatial pressure !radient is
called convection. nalo!ously, system equilibrium can be broen by modifyin! (increasin!
concentration, for e"ample) the chemical potentiala
i of only one component in phase a
(component i) without alterin! all other parameters (temperature, pressure and remainin!
components chemical potentials). !ain, amon! the possible ways the system has to !et a
new equilibrium condition, the simplest one is to rendera
i, equal tob
i, without alterin! all
other parameters. 3hemical potential re-equilibration implies the mo%ement of component i
from phase a to phase bcrossin! the permeable membrane untila
i, +b
i, . In this case,
howe%er, mass transport is the result of the random, thermally induced, rownian motion of i
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molecules as no pressure !radient e"ists to determine a con%ecti%e flow. Mass flow induced
by a chemical potential !radient is called diffusion.
1.3.1 Mass conservation law
ccordin! to the mass conser%ation law, mass can be neither created nor destroyed so that, in
a fi"ed space re!ion, mass can increase or decrease only by addition or subtraction from the
surroundin!s, respecti%ely. lthou!h this law can not be applied in the case of radioisotope
decay or nuclear fission where mass con%erts into ener!y accordin! to the well-nown
5instein equation, in the typical pharmaceutical and en!ineerin! 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 !eneric flowin! fluid composed by rcomponents. In the abo%e mentioned fi"ed space
re!ion (that, for the sae of simplicity, can be thou!ht as a parallelepiped, see 'i!. 4.#), the
conser%ation law translates into a mass balance that, for the !eneric ithfluid component can be
e"pressed as6
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=
reactionby
lossor!ainofrate
ssurrondin!from
lossofrate
ssurrondin!from
!ainofrate
chan!e
massofrate iiii (4.#7)
where the last term (calledgenerative term) refers to a !eneric reaction that can, in principle,
in%ol%e component ith and can cause i loss or !ain dependin! on whether i is a reaction
reactant or a product. 8b%iously, to sol%e the problem of mass transport, (r- $) eq.(4.#7) lie
equations plus an o%erall mass balance must be written6
#
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=
ssurrondin!from
lossmasstotalofrate
ssurrondin!from
!ainmasstotalofrate
chan!e
masstotalofrate (4.4$)
where, now, the !enerati%e term disappears as the chemical reaction can comport neither a net
mass !ain, nor a net mass loss but only a fluid composition chan!e. If X, Y and Z
represent our parallelepiped dimensions in the X, Y and Z 3artesian a"es directions,
respecti%ely, the first term of eq.(4.40) will be !i%en by6
ZYXt
Ci
=
i
chan!e
massofrate(4.42)
where t is time, Ci is ith component concentration (mass9%olume) in the %olume V +
X*Y:Z. /he second term in eq.(4.#7) can be computed rememberin! that the incomin!
mass flow is !i%en by the sum of mass normally (i.e. in the direction of surface normal)
enterin! the three surfaces S;
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(area + X:Y and positioned inZ+ Z normalZa"is, see 'i!. 4.#). ccordin!ly, if J=, J;
andJ
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i
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=
X
C
X
C!vCJ
G
Gii
im=i=
=Y
C
Y
C!vCJ
G
Gii
im;i;
=
ZC
ZC!vCJ G
Gii
im
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=
=r
i
C$
i G (4.@#)
0E$
iim ==
r
i
! (4.@4)
=
=r
i
G$
i 0 (4.@@)
5q.(4.@#) states that fluid density is the sum of each component concentration, eq.(4.@4)
affirms that no net fluid motion can tae place due to diffusion while eq.(4.@@) maes clear
that fluid mass can not modify because of a whate%er chemical reaction. ccordin!ly, the
well-nown continuity e'uationis found6
( )vGG =t
(4.@A)
Hotably, this equation can be also deri%ed startin! from an o%erall mass balance similar to
that e"pressed by eq.(4.4A) assumin!, Gi+ 0, =i= vCJ = , ;i; vCJ = and
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where )i is the net char!e on species i, ( is 'araday constant and is the electrostatic
potential. tartin! from this definition, after some manipulations, it is possible to deri%e the
well-nown Herst-lan equation6
Niiiii =$T
(C)C!J (4.@F)
where Ji is the ith component flu", $ and T are, respecti%ely, uni%ersal !as constant and
absolute temperature. It is clear that the first and second terms on the ri!ht represent,
respecti%ely, the fician and the electrostatic potential !radient contribute to diffusion.
1.3.2 Momentum conservation law
lthou!h rinetics equations are now a%ailable, the problem of !ettin! components profile
concentration in the flowin! fluid is not yet sol%ed as the nowled!e of the fluid %elocity
field is required. Indeed, eq.(4.4A&) maes clear (see the term ( )viC ) that the time and
space %ariation of the fluid %elocity %ector v must be nown to !et Ci space and time
dependence i. /o sol%e this problem, recourse must be made to a particular force balance.
It is well nown that the %elocity vof a body of mass m (for the sae of simplicity here
thou!ht of spherical shape) chan!es with time due to the action of all the forces actin! on it
accordin! to Hewton&s law6
== id
dF
va
tmm (4.@7)
where ais body acceleration and Fiis the !eneric force actin! on the body. In these terms, eq.
(4.@7) holds if mis a constant. Dhen, on the contrary, mis time dependent, eq.(4.@7) has to
be !eneralised. part from relati%istic effects tain! place near li!ht speed, mcan be time
dependent also in much more common situations. t this purpose, let&s focus the attention on
a cubic %essel containin! an amountof sand and slidin! (see 'i!. 4.4) on an inclined plane.
If at time t0a hole is opened on the %essel bac face, sand will pro!ressi%ely !o out and
decreases with time causin! a system mass m(sand plus %essel mass) decrease. In order to
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describe this situation, eq.(4.@7) must be !eneralised assertin! that the %ector p+ mv, called
linearmomentum orsim%ly momentum, chan!es because of the action of all the forces actin!
on the %essel and because of sand lea%in! from the hole. ccordin!ly, subOect of our attention
is no lon!er body %elocity (v) but body momentum (p+ mv). /his scenario can be further
complicated ima!inin! that %essel motion taes place under a sand rain. ccordin!ly, if on
one hand, mtends to decrease for sand escapin! from the hole, on the other hand, mtends to
increase due to sand comin! in throu!h the upper open %essel surface (see 'i!.4.4). In
conclusion, the time %ariation of system momentum ( t
m
t d
d
d
d vp
= ) depends on the sum of
forces actin! on the body plus the incomin! and escapin! momentum due to in- and out- sand
flow rates, respecti%ely. /his simple e"ample can be used to understand what happens to a
fluid element (that, a!ain, can be thou!ht as a parallelepiped of %olume Vand dimensions
X, Yand Z) of a flowin! fluid. Indeed, as in the %essel e"ample, the %ariation of the fluid
element momentum (t
V vG ) is due to the action of all the forces actin! on it, namely
surface and body forces, plus the difference between the incomin! and e"itin! momentum
across parallelepiped surfaces. If the e"pression of surface and body forces is not so
complicated, as shown later, much more care must be de%oted to the e"pression of the inflow
and outflow momentum. Indeed, as momentum is a %ector and it is characterised by three
components, it is con%enient, for the time bein!, to limit our considerations to its X
component per unit %olume (the scalar quantity v= can be thou!ht as the momentum X
component concentration per unit %olume), bein! the followin! ar!umentations e"tensible to
the other remainin! two components, namely Y (v;) and Z (v
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and not a %ectorial one (v). ccordin!ly, as in to the case of the mass conser%ation law
(para!raph 1.3.1), we are a!ain dealin! with the e"pression of a scalar quantity (v=) flu"
throu!h a parallelepiped. 3onsequently, the net balance throu!h the two surfaces (S;X, reads6
( )[ ] ( )[ ]( )=======
GG + vvvvZY (4.A0)
imilarly, the balances throu!h the two surfaces (S=Yand the two surfaces (S=;) of normalZand placed inZandZ>Zread, respecti%ely6
( )[ ] ( )[ ]( );;;=;;= GG + vvvvZX (4.A$)
( )[ ] ( )[ ]( )
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#i, , this meanin! that is associated e"clusi%ely to fluid motion. In addition, it can be
demonstrated Melt1 that is symmetric, this reflectin! in the followin! relation iO+ Oifor iQ
#. In order to be homo!eneous with the strate!y adopted in writin! eqs.(4.A0)-(4.A2), also in
this case only theXcomponent of the three stress %ectors will be considered (namely, ==, ;=
and
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( ) ( ) ( ) ( ) ;
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( ) ( ) gv GR
RG += P
t (4.C0&)
Hotably, this equation is formally equal to eq.(4.@7) as it states that the product of mass (per
unit %olume) and acceleration is equal to the sum of all the forces actin! on the fluid element.
In other words, eq.(4.@7) structure can be maintained also for the description of a fluid
element motion pro%ided that traditional acceleration is substituted by a substanti%eJ
acceleration.
In order to render eq.(4.C0) operati%e, it is necessary to define the relation between the stress
tensor components and the %elocity !radient. 'or a newtonian fluid, in 3artesian coordinates
(see chapter #), we ha%e6
( )v
+
= -S#
2S2P ===
X
v(4.C#)
( )v
+
= -S#
2S2P ;;;
Y
v(4.C4)
( )v
+
= -S#2
S2P