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In the last lecture, we saw equations of fluid motion namely the law of mass conservation
applied to the bulk fluid as well as the second law of motion also called the Navier-Stokes
equations. Moving fluids also carry with them scalar
quantities. Today, we are going to look at equations that govern transport of scalar
quantities.
The first scalar quantity is the specie in a mixture. For example, if you are dealing
with a combustion problem then the species carried are oxygen, carbon dioxide, fuel,
carbon monoxide and so on and so forth. The law of mass conservation for a specie
in a mixture is called the mass transfer equation. We will also invoke the first law of thermodynamics
which transports energy in a moving fluid. Both energy and the concentration of the specie
are scalar quantities.
What does the law say for a specie? For a specie k in a mixture, the rate of accumulation
of mass of the specie k that is given by M dot k accumulation equals rate of mass in
minus rate of mass out plus rate of generation of specie within the control volume.
As a result of chemical reaction, as you know some species are generated and while others
are destroyed. As such each specie will have either a generation or a destruction rate
associated with it. What is M dot k accumulation? That is simply the mass of the specie k within
the control volume delta v - d by dt of rho k delta v.
The mass of specie in will be the flux of specie k in direction 1 multiplied by the
area d A 1 which is this area plus N 2 k which is coming from the bottom multiplied by the
area delta A 2 plus N 3 k multiplied by delta A 3 at x 3 and likewise the same quantities
at the outgoing phases at x1 plus delta x 1, x 2 plus delta x 2 and x 3 plus delta x
3; rho k is the specie density. If we now substitute these expressions in
this verbal statement and divide each term by delta v which is the product of delta x
1, delta x 2 and delta x 3 and let each of these increments tend to 0; this is a procedure
we have gone through before.
Then you will see you will get an equation of this type d rho k by dt plus net transport
of specie k in direction 1, plus net transport in direction 2, plus net transport in direction
3 equal to rate of generation of specie k. Now, the total mass flux N i k in direction
i is the sum of the convective flux due to rho k u i due to bulk fluid motion and diffusion
flux m double prime i k due to density difference. Thus N i k is represented as rho k u i plus
m double dot i k. Writing convective flux in this manner is
indicative of the fact that we are assuming that each specie is traveling or is being
carried at the same velocity as the bulk fluid. The diffusion flux on the other hand m double
dot i k arises simply due to the differences in density at neighbouring locations.
The expression for mass flux by diffusion is given as m double prime i k equal to minus
D rho k by d x i. This equation is analogous to the conduction heat transfer and is like
a Fourier's law of heat conduction. In mass transfer, it is called the fix law
of mass diffusion. So, just as in convective heat transfer you know that the total flux
of energy is given by the convecting flux plus conduction flux. In mass transfer, we
say it is the convective flux plus diffusion flux.
D is called the mass-diffusivity; it has units of meter square per second. So, if I have
to substitute for N i k for each of these terms then I can rewrite this equation in
the following manner.
It would read as d rho k by dt plus d rho k u 1 by d x 1 and so on and so forth is equal
to d by d x 1of d rho k by d x 1 and likewise in direction 2 and 3 plus R k.
Rho k has units of density. It is customary to define mass fraction omega k as the specie
density divided by the mixture density and therefore, sum sigma omega k equals 1 by Dalton's
law. Another way of saying is sum of densities is equal to the mixture density.
The notion behind Dalton's law is that each specie behaves as though it occupies the volume
of the total mixture. If I were to substitute now, rho k as rho
m multiplied by omega k, this equation would be written in tensor notation in this fashion
this is called the mass transfer equation. It has a transient term, a convection term,
a diffusion term and a source or a generation term.
Now, let us sum each of these terms over all species - that is sum over all k as done here.
Then clearly here this term would reduce to d rho m by d t; this term would reduce to
d rho m u j by d x j because omega k is equal to 1.
Sigma omega k will simply become unity or constant. Therefore, this term would simply
vanish. Just see now that this equation would be d rho m by d t plus d rho m u j by d x
j equal to sigma R k and that term is here. You will recognize readily then in the absence
of omega k, the left hand side is simply the bulk mass conservation and it equals 0; it
follows therefore the sigma R k must be 0. These two last deductions are very important.
sigma R k equal to 0 says that whenever there is a chemical reaction it is true that some
species will be generated, but there would be others that would be destroyed and the
total mass cannot be generated or destroyed; that idea is expressed in sigma R k equal
to 0.
What does summation of diffusion equal to 0 imply? It implies that when some species
are being diffused in a certain direction, other species are being diffused in the opposite
direction and that stands to reason. For example, if fuel was decreasing then product
would increase. The net result is that summation of all diffusive quantities would be 0 and
therefore, sigma N i k would be sigma rho m u i plus sigma m double prime i k, but that
quantity is 0 and therefore, sum of the mass fluxes in direction i are simply the bulk
mass flux rho m u i. We shall refer to this equation much later in the course.
A word about Mass Diffusivity: strictly speaking mass diffusivity is defined only for a Binary
Mixture of two fluids 1 and 2 as D 1 2. But in a combusting product mixture for example,
there are several species present and diffusivities of pairs of species are truly different. So,
diffusion of carbon dioxide in nitrogen or the diffusivity of carbon dioxide in nitrogen
is different from diffusivity of oxygen in nitrogen and vice versa, but in gaseous mixtures
these diffusivity pairs between species tend to be very nearly equal and therefore, we
discard D 1 2 or d i j as the symbol for diffusivity and replace it by single symbol D and that
suffices for combustion calculations. Incidentally in turbulent flow, this assumption
of equal effective diffusivity holds even greater validity; this we shall appreciate
a little later.
We now turn to the first law of thermodynamics in rate form that is Watts per cubic meter.
The first law of thermodynamics reads as d E by d t equal to d Q convection by d t plus
d Q conduction by d t plus rate of generation of energy minus work done by shear forces
minus work done by body forces. Each of these terms is defined here and we
shall seek mathematical representations for each one of them.
So, let us consider the first term E dot; that would simply be d by d t rho m e naught
or the total energy. The total energy is the sum of static energy plus kinetic energy and
the static energy by a thermodynamic relation is nothing, but enthalpy minus p into specific
volume or p divided by rho m plus V square by 2.
So, e m is mixture specific energy, h m is mixture specific enthalpy, V square is the
kinetic energy - sum of u 1 square, u 2 square and u 3 square and rho m as we saw before
is sigma rho k, the Mixture Density. In general, e naught will have contributions
from many other force effects like potential energy associated with rise or fall in elevation,
electro-magnetic energies and so on and so forth, but these we will neglect because practical
equipment's are fairly small. But in which Yes, kinetic energy would be
of some interest, but not these energies that I mentioned at the moment.
Now, in order to represent the heat transfer and the work transfer across a control volume
we shall follow the thermodynamic conventions. We will say that heat flow into the control
volume is positive whereas, the heat flow out of the control volume is negative.
As we said before, all species are transported at mixture velocity. Here, I have shown all
the net mass transfers and heat transfers that could take place across a control volume
phase. N is total mass flux and q is the total heat flux across the control volume phase.
Net convection then by following the convention that the heat in is positive whereas, heat
out is negative, we would have convection as simply d by d x j of sigma N j k e naught
k and if I replace e naught k by h k minus p by rho m plus V square by 2.
Then you will see N j k would multiply with h k into the bracket, but the sum of N j k
would simply be rho m u j as we saw before minus p by rho m plus V square by 2. If we
now note that omega k into h k, that is the mass fraction of specie k multiplied by its
specific enthalpy must add up to the mixture enthalpy; then sigma N j k h k would be sigma
rho m u j omega k plus m double dot j k, the diffusion flux into h k.
Since sigma omega k h k is equal to h m, it will add up to rho m u j h m plus sigma m
double prime j k h k. Therefore, if I replace this term here then
you will see you have rho m u j h m minus p by rho m plus V square by 2 which would
be written in this fashion and then the left over term, the diffusion flux term that could
be given by that term. So, that is the expression for the convective flux.
Net conduction again by Fourier's law of conduction: Q conduction will be simply minus d q j by
d x j where q j is k m multiplied by d T by d x j and k m is the mixture conductivity.
Net volumetric generation: Now, volumetric generation in a moving fluid typically would
comprise of the generation due to chemical energy because some reactions are exothermic
whereas, some other chemical reactions are endothermic. So, Q dot chem would be positive
in case of exothermic reactions and it would be negative in terms of endothermic reaction.
If you want to evaluate Q dot chem in a moving fluid then one needs to postulate first of
all the chemical reaction model that is being employed and there are variety of chemical
reaction mechanisms of different levels of complexity. We shall see all these when we
come to study of mass transfer. Q rad represents the net radiation exchange
between the control volume and its surroundings. Usually evaluation of this term would require
what is called the radiation transfer equation and it happens to be an integro-differential
equation.
Treatment of that equation is really beyond the scope of the present lectures and it would
require usually numerical calculations in a real practical equipment, but suppose the
mixture that we are carrying has very high absorptivity may be because of soot, may be
because of particulates that are present in the gaseous mixture.
Then the absorptivity and scattering of the radiation would be very high. When absorptivity
and scattering coefficients are large, Q dot rad can actually be represented in a manner
similar to the conduction equation. The radiation conductivity can be defined as 16 into sigma
T cube divided by a plus s where sigma is the Stefan-Boltzmann constant that you are
all familiar with. But remember this kind of representation is
justified only when the gaseous mixture has very high absorption and scattering coefficients.
Now, we come to the work done terms and the derivation to follow is somewhat longish and
I would request you to play good attention to how the derivation progresses.
Firstly there are shear forces and normal forces. Here is the stress; stress is force
per unit area multiplied by velocity gives you the work done and d by d x 1 of that gives
you the net work done. So, sigma 1 which acts in the direction 1
multiplied by u 1 plus tau 1 2 into u 2 is the work is the shear work in net shear work
in x 1 direction and so on and so forth. You will get shear stress multiplied by the
associated velocity in the two directions and the net work done by stresses W dot s
would be given as that. The body forces work is rho m into B 1, the
force multiplied by the velocity u 1 is the work done in direction 1, 2 and 3 likewise.
So, W dot s is the stress work; W dot B is the Body-Force Work.
Now we shall write these things as differentiation of a product. So, this d by d x 1 sigma 1
u 1 will be written as sigma 1 d u 1 by d x 1 plus u 1 into d sigma 1 by d x 1 and so
on and so forth.
On the next line you will see the result. So, the total work-done will be u 1 multiplied
by all these terms, u 2 multiplied by all these terms, u 3 multiplied by all these terms,
plus sigma 1 d u 1 by d x 1, sigma 2 d u 2 by d x 2, sigma 3 d u 3 by d x 3, tau 1 2
into the strain rate associated with tau 1 2, tau 1 3 into the strain rate associated
with tau 1 3, and tau 2 3 into the strain rate associated with tau 2 3 and here I have
already used the idea of complementarities of stresses that is tau 1 2 is equal to tau
2 1. Now I draw your attention to this term. If you recall, when we wrote the Navier-Stoke's
equations before or the Newton's second law of motion, these are simply the net forces
acting in direction 1 and therefore, these are nothing, but the right hand sides of the
Newton's second law of motion. They can be replaced by the left hand side
of the Newton's second law of motion and that is what I will do on the next slide plus all
the stresses can be replaced in terms of viscosity multiplied by strain rates.
Let us do this first. So, multipliers of u 1, u 2, and u 3 in equations 18, 19 are simply
right hand sides of momentum equations, if you recall that on lectures 3 slides 13-14-15.
We have replaced them by left hand side of momentum equations. Left hand side was u 1
D u 1 by D t into rho m u 2 D u 2 by D t and u 3 D u 3 by D t.
This would simply be D u 1 square by 2 D t, D u 2 square by D t and D u 3 square by D
t which we write as D V square by 2 D by D t into rho m. These are the first 3 terms
that we have expressed; now we look at the remaining terms.
If I replace sigma 1 by minus p plus tau 1 1, sigma 2 as minus p plus sigma plus tau
2 2 and sigma 3 as minus p plus tau 3 3 and tau 1 2 as mu times D u 1 by D x 2 into D
u 2 by D x 2 and so on and so forth.
You will see that the term would be equations 21 and 22 would be simply mu phi v minus p
del dot V where phi v is called the Viscous Dissipation Function.
It would take the form 2 times d u 1 by d x 1 square which arises from tau 1 1, d u
2 by d x 2 square, d u 3 by d x 3 square and so on and so forth and del dot V as we recall
is simply, d u 1 by d x 1 plus d u 2 by d x 2 plus d u 3 by d x 3.
Therefore, the total work done terms are simply rho m into D by D t V square by 2 plus mu
phi v minus p del dot V. You will notice that phi v will always be positive because it is
a sum of all the gradient square. So, that sum would always be positive and
in general and its purpose is to increase the energy level of the mixture. This is the
kinetic energy term and this is the minus p d V term or sometimes also called the pressure
work term. Del dot v as you know is this.
By way of summary: we wrote expression for E dot as d rho m e naught by d t; expression
for convection was written as minus d rho m u j e naught by d x j minus diffusion heat
transfer d by d x j of sum of m dot j k h k.
Q dot conduction was written in this fashion; Q gen was Q dot chem plus Q dot rad and lastly
we saw this term is this kinetic energy term, viscous dissipation term and so on and so
forth. You will see that mu phi v is positive and
therefore, it tends to increase the rate of energy with time, conduction and if I were
to replace Q dot rad as k rad d T d x j and then it would simply get added to that term.
This is the p del dot V; this can be both positive and negative and so can this be positive
and negative.
The final energy equation then reads as follows. What I have done is simply transported this
negative term to the left hand side so that it becomes positive and we have d rho m e
naught by d t plus d rho m u j e naught by d x j equal to net conduction heat transfer.
Net conduction heat transfer due to mass diffusion, net kinetic energy generation minus p del
dot V plus Viscous Dissipation, chemical energy generation and radiation generation.
Now, if I replace e naught as h m minus p by rho m plus v square by 2 then this term
would become simply rho m D h by D t. The kinetic energy term on the left hand side
would cancel with this kinetic energy term and the differential of rho m u j p by rho
m which is essentially u j p would simply give me D p by d would cancel partly with
minus p del dot V and I will have a term called D p by D t plus Q dot chem rad and so, this
is called the Energy Equation written in enthalpy form.
This is the energy equation written in internal energy form. It is this equation which is
largely used in flows with or without chemical reaction.
Notice that all it says is that rate of change of enthalpy is equal to the net conduction
heat transfer, net mass diffusion heat transfer and total viscous dissipation.
D p by D t - this term is first of all the D p by D t will be partial D p by partial
D t which would be of importance in unsteady flows such as one in which explosions are
studied plus u into D p by u 1 into D p by d x 1, u 2 into D p by d x 2 plus u 3 into
D p by d x 3 - these terms are important particularly when shocks are present where very large pressure
gradients occur in a given direction, but in low speed force really this term is relatively
unimportant and we usually tend to ignore it.
Q dot chem would arise out of the combustion model that one I specified and radiation is
to be included in very high temperature reacting flows.
I will conclude here our equations of energy. I would leave you with the simply recall of
the mass transfer equation that we derived that was this d rho m omega k by d t and so
on and so forth. This is the mass transport equation and then the last one is the energy
transport equation. It is these equations that we shall develop further.