In our last few lectures, we have derived
the governing equations of fluid flow and
we have derived this specifically for the
general case where there may be chemical reactions
that may be happening and there may be heat
transfer and mass transfer that may be happening
in this and in the within the fluid continuum.
So, in today’s lecture which will take as
a tutorial, we will just take a deep breath
and then try to write down all the equations
and then see what kind of equations that are
there for us to solve numerically. So, we
will write down all the governing equations
that we had derived in the last few classes,
and these equations comprise the mass balance
equation, the momentum balance equation, the
energy balance equation and in the case with
chemical reactions we also have to have the
species balance equations.
In all the lectures that we have been dealing
with, we have started out with the Cartesian
coordinate system and somewhere at some point
we have also put down the equations in vector
form. In today’s class, we will start with
the vector form and then so that it is easy
and compact for us to write them down, and
if you want you can go back to the previous
lectures and then write the equivalent Cartesian
coordinate system form for easier understanding.
So, we will start with the equations governing
fluid flow.
We are considering the case of a reacting
fluid in which you have a homogeneous reaction
which is given in a stoichiometric way as
like this, we have a r 1 j X 1 plus a r 2
j X 2 plus so on these are the coefficients
stoichiometric coefficients of reactant 1,
reactant 2, reactant 3 and so on in the jth
reaction. So, we are looking at a large number
of this reactions which may be happening in
which species X 1, X 2, X 3 and so on are
participating as reactants and these are getting
converted to products a p 1 j X 1 plus a p
2 j X 2 and so on here.
So, in a given system of chemical reactions
having N S number of species, participating
in N R number of reactions, in this sometimes
some of the reactants may appear as reactants
and sometimes they may appear as products.
So, this is a general scheme here and in this
we are assuming that the kinetics of the reaction
scheme 
are known, all the kinetic constants in terms
of the, along with the order of the stoichiometric
coefficients as here, along with that we need
we already know the order of each reaction
and then the constants involved in the rate
expressions all those things are known and
they are part of the problem’s specification
here.
So, if you are now looking at fluid flow,
a single phase fluid flow in which consists
of a mixture of all these species which are
participating in this, then as the flow is
taking place as heat transfer and mass transfer
all those things are taking place, there are
the fundamental laws of mass conservation,
momentum conservation, energy conservation,
equation energy conservation and species conservation
equation have to be obeyed and the corresponding
laws are what we are going to write down here
and we will start with the mixture mass conservation
equation.
And we are writing them down for a constant
property fluid, we can always write down the
more complicated form, but for our easy understanding
we will take down the simplest form with constant
properties 
of the fluid including the species. So, all
the relations like thermal conductivity, specific
heat and other properties of individual species
and of a mixture are expected to be known
and constant when we are writing down this
simplified form of equations, and here, if
rho is the density of the mixture, which is
expressible as sum of the densities of the
individual species.
We can we can write down the mixture equation
in this particular form, where rho is the
density of the mixture mass density of the
mixture and u is the mass velocity of the
of the mixture and we also have the momentum
conservation equation for the mixture is again
dou rho by dou t. In fact, here rho is constant.
So, we can cross this out and then you can
take it out and that becomes del dot u equal
to 0 and here we have dou u by dou t plus
del dot uu, this is the dyadic velocity vector
tensor is given by nu del square u plus g.
So, we can write this also as dou u by dou
t plus del dot del u plus g here. So, this
is the momentum conservation equation. Thank
you and we have also forgotten the pressure
gradient so that comes out here like this.
Minus 1 by rho gradient of pressure, we now
have the energy equation; energy conservation
equation. This again can be different types
of energy and we have finally, derived an
expression for in terms of enthalpy from that
assuming constant properties, we had an equation
written in terms of temperature. Okay, here
we have more terms. We have dou T by t with
capital T is a temperature plus del dot uT
and T is of course, a scalar whereas, u is
vector here and del is the also a gradient
operator and it is a vector and that is equal
to k by rho cp times del dot del T, we have
just put the del square T as del dot del T
plus mu by rho cp times phi v, this is all
the rate of work done by the viscous forces
and Q R is the heat release because of the
chemical reaction that happens in this divided
by rho cp and in this equation we have neglected
any contribution from radiative of heat transfer.
So, this is there is no radiative heat transfer
involved in this if there is a radiative of
heat transfer and if there is a flux of radiative
heat coming from the surfaces of the control
volume then that will also come as additional
heat flux term through the control surfaces
okay. So, this is the energy conservation
equation 
and we also need to have a definition of density
and all that. In fact, for a mixture we need
to keep these things.
So, we need to have the species balance equation
and this we have put in the form of the mass
fraction of the species.
So, we can we have put this in the form of
species mass fraction Yi as dou by dou t of
rho Yi plus del dot rho u Yi equal to del
dot gamma gradient of Y plus S of i and what
the subscript i here indicates is a species,
for example, Ya, Yb, Yc and so on or Y oxygen
or y carbon dioxide, Y steam those type of
things and so this is a mass fraction of the
species and something come coming in and going
out because of the flow and diffusing and
S is the essentially the rate of formation
and we need to have an expression for the
rate of formation of this species here. We
need to have an expression for the rate of
formation of this species here and similarly
we need to have an expression for the rate
of heat released because of the chemical reaction
and both of these depend on the reaction mechanism
here.
And we can write down the values of the source
term, which is going to appear in that in
terms of the molecular weight and the individual
rates of the reaction, which is given as okay,
so this is the rate of reaction of the jth
reaction and in this reaction the ith reactant
is appearing as a reactant with this stoichiometric
coefficient in the jth reaction and therefore,
it is disappearing at this rate, and it may
also be appearing as a product with the coefficient
of a p i j. So, in which case it is being
formed so that is an addition and the rate
of formation is given by the rate of that
particular reaction and the stoichiometric
coefficient here and in a general case you
have either this or this as non-zero or both
as 0. So, it is unusual to have both being
non-zero in the same reaction.
So, when it appears as a reactant the source
term is negative and when it appears as a
product the source term is positive, and the
rate at which it happens is given by the stoichiometric
coefficient of participation as a reactant
or as a product times the rate of reaction
and this can happen in any or all of the j
reactions. So, we have to sum over all the
reactions, all the j number of reactions in
order to find out the total rate of formation
of this and here these are in terms of stoichiometric
coefficients.
So, we need to get that into multiply by the
molecular weight of this species in order
to get the source rate of formation here and
similarly the 
heat release is expressed in terms of; this
is sum over all the reactions, sum of the
product of the rate of reaction, okay and
so 
what we have here is in a particular jth reaction,
ith species is being formed. So, you have
that delta H, heat of formation of the ith
species and it may be disappearing because
it is being consumed as a reactant here. So,
that is minus here and it may be appearing
as a product in which case it is a plus. So,
whenever you, it is product of is formed then
this heat of formation becomes positive quantity.
So, this happens for all the species, this
has to be summed over all the species that
are participating in this reaction and times
the rate of that particular reaction and that
is summed over all the reactions here will
give us the total heat of the, this system
of reactions. So, we have these rates of reaction
and the heat of reaction terms which appear
in the energy balance equation and the species
balance equation okay. So, we have to be careful
about the units and I leave it as an exercise
for you to see that; see give the correct
units for Q R and S i, every equation of this
has to be dimensionally consistent.
So, that is the sum total of these units of
this particular term must be same as the units
of this term and similarly the units of this
and this and all those things must be the
same okay. So, now, these are the 4 equations
that we have put in blocks here are the equations
that govern the flow of this reacting fluid
containing N S number of species, which are
reacting as per this N R number of reaction
scheme which is given by the reaction scheme
here and the rates of the reaction.
Now, how many equations are there? We have
a total of 1 continuity equation plus 3 momentum
equations plus 1 energy equation plus N S
minus 1 number of species balance equations.
So, this is species, this is energy and this
is momentum 
and mass balance. So, that is N S plus 4 equations
are there and what are the variables. So,
these are the equations, we have the mixture
u, v, w, the mixture pressure, the mixture
temperature and the mass fractions of the
species and there are N S number of species
okay. So, we have , so this is N S number
of species, but we know that in addition to
this we have sigma Yi equal to 1, so, we need
to know only N S minus 1 number of these to
get the last one here. So, we have N S minus
1 number here plus 5. So, the total number
of unknowns is N S plus 4. So, this total
number of unknowns is N S plus 4 and the number
of equations that are available is also N
S plus 4. So, these two are the same.
So, we have these 4 equations plus the condition
since we have put the species balance equation
in terms of the mass fraction all the species
will have to add up to a total of 1 is an
extra algebraic constraint that is coming
out and because of that we need to know only
N S minus 1 number of Yi’s and the last
one, the remaining one is known from this
equation. So, here we have, so many number
of equations, in the simplest case of no chemical
reaction and a single component fluid like
that then the same equations apply here this
becomes redundant and rho becomes the density
of the fluid and nu becomes the viscosity
here and in this equation Q R will be 0, but
all the other terms will remain.
So, you can have Q R becomes 0 and S i becomes
0 and Yi is equal to 1. So, it is nothing,
it is essentially the same as this one here
because if Yi equal to 1 gradient of 1 is
0. So, this term is 0 and here this becomes
this and this becomes this. So, the species
conservation equation reduces to the mass
conservation equation which is not surprising
and you have the other equations remain the
same except for the heat of reaction which
appears in the energy equation.
So, when you look at these equations, these
equations have a characteristic form, which
is both advantageous for us and it is also
something that we have to take note of when
we attempt a solution of these and that characteristic
form is what is essentially known as the transport
equation form where something is written in
terms of dou phi by dou t plus del dot u phi
equal to del dot gamma phi times gradient
of phi plus source term here. So, this is
the standard form of this equation, the rate
of variation with respect to time of phi at
a particular point x y z within the whole
domain or within the fluid continuum is given
in terms of, what is a term, which is known
as the advection term here, the change in
a particular at a particular point may because
of the, may arise because of the convective
nature of the flow because of the flow because
fluid velocities the quantity phi may change.
This is known as the advection.
So, that is the net difference between what
is being taken out and what is being brought
in into our control volume and that is the
advection term. So, associated with the flow
of the particular fluid and this is the diffusion
term, counter gradient diffusion is what is
actually bringing in either momentum okay
here and heat and also the mass transfer here,
that may be happening and so this is the diffusion
of phi because of gradients of phi within
the fluid continuum is what is causing this
and this is a source term okay. And this source
term takes different forms, for example, here
in the momentum balance equation pressure
gradient is a source term and gravitational
external force is a source term.
In the case of energy equation, energy conservation
equation the viscous dissipation term here
is a source term which will be leading to
increase in temperature because of the flow,
because of the work done again viscosity,
if you have an exothermic reaction than that
heat of the reaction may be a source term
and in the species conservation equation,
the rate of formation of the species is a
source term. It can also be a sink term if
it is negative. So, these all these equations
can be put in this particular form, variation
of a quantity is expressed in terms of the
advection term, diffusion term and a source
term with a corresponding velocity which is
nothing but the fluid velocity. A diffusivity
which varies in the case of in this particular
case with written, when phi is equal to u
here this is the kinematic viscosity and phi
for the energy equation is k by rho cp and
here it is the diffusivity with a proper dimensions
okay and the source term for phi can also
be different for different equations, for
the continuity equation it is 0 and here it
is as mentioned the gravity term and the pressure
gradient and these things like this.
So, this is a general form of the equation
and this particular form has 2 important contributions;
one is the advection okay which means that
it is directional and it is associated with
the flow direction and diffusion which happens
in all directions. So, part of it is direction,
directional and part of it is essentially
adirectional it is all directions not necessarily
the same. If it is isotropic then it is the
diffusivity is same in all directions sometimes
it is not necessarily isotropic, in which
case this can happen more in a certain direction
and less in a certain direction as a result
of both these processes you have a variation
of phi within our domain x, y, z as a function
of time will be changing as per this equation.
So, in the next lecture, we will be looking
at having looked at all this partial differential
equations which govern the fluid, we look
at what are the appropriate boundary conditions
given this specific nature of the governing
equations which is where the quantity that
is of interest is governed partly by directional
terms like advection and adirectional terms
like the diffusion term and source term can
be different they can be independent of the
parameter phi that we are looking at or in
some cases they can also be functions of the
parameter.
So, in, we will be looking at the boundary
conditions and the initial conditions that
are required to complete the mathematical
formulation of the problem, just the governing
equations are not sufficient along with that
we need to have the boundary conditions. So,
we will discuss the boundary conditions and
initial conditions which are necessary for
us to have a well posed problem. That is a
problem, which has a potentially unique solution
and it is the unique solution that we wish
to find through a numerical solution. So,
that would complete the next this particular
module and we will deal this with this in
the next class.
Thank you.
