In the last lecture, we derived the energy
equation. And the set of equations which govern
incompressible constant property flow with
heat transfer is given here in this slide.
We have the continuity equation, the three
momentum conservation equations for a Newtonian
fluid with constant properties, so that the
viscosity - the kinematic viscosity here is
constant; and the energy conservation equation
reduces to a much simpler form and expressed
in the form of a rho C p T that is specific
heat, constant pressure. And therefore, from
in order to get to this particular form, we
have rewritten the earlier equation which
we had written in terms of internal energy
that has been rewritten in involving a enthalpy.
If we do that process, you will get an equation
like this.
And as already mentioned, this is viscous
dissipation term and when it is negligible
unless you are dealing with large viscosity
fluids large viscosity liquids. For example,
if you are studying the mixing of a highly
viscous fluid using an impeller, at that point,
we would have to consider this because this
would generate a sufficient amount of local
dissipation and local increase in temperature
which may have a consequent large change in
the viscosity which would then be coming into
the momentum equation through this term and
that could change the velocity field.
So, in such cases, you would have to consider
this, but if you are looking at light fluids,
light viscosity fluids liquids and gases,
then you could neglect this particular term
and you could take this as essentially the
thermal energy conservation equation. And
solution of this will give us temperature
as a function of x, y, z; solution of these
will give us u, v, w and p as a function of
x, y, z. And from these, we can get the friction
factors or pressure drops, the heat transfer
coefficients, the amount of heat transferred,
the wall temperatures, all this information
we can get from this.
So, what we will do now is that using these
equations, we will see heat transfer applications,
and how we can relate these equations to our
conventional non CFD conception of heat transfer.
So, we are looking at heat transfer applications
involving the three modes of heat transfer
- convective heat transfer, conductive heat
transfer and radiative heat transfer. This
radiative heat transfer especially from a
point of view CFD is quite complicated, so
it is not relay in the syllabus as well as
the exams is concerned, but I do want to spend
a few minutes discussing this, so that you
know that there is much more to it then what
we have seen so far in this.
So, that it is not something that is included
in what we have doing. So, to that extent,
this more an awareness it is not going to
be there from the point view of exams. So,
the idea is that having formulated a problem
in terms of having a derive the governing
equations, how can we establish that what
we have done is capable of dealing with convective
heat transfer and conductive heat transfer
and radiative heat transfer, so that is a
question that we would like to examine now.
So, when we look at convective heat transfer.
So, when heat transfer is to be included,
the energy equation has to be solved along
with the Navier stokes equations. Because
it is only from the energy equation, we get
temperature and only when we have temperature
we can talk about heat transfer. So, but we
also see that the temperature equation has
a velocity component that is coming here and
there is also viscous dissipation.
So, even if we neglect this we still have
a velocity component that is coming here;
and in that sense the velocity field also
needs to be resolved and that means, that
we have to solve the energy equation together
with the momentum equation and the continuity
equation. So, here we have one form of energy
equation written in terms of the total energy
which is internal energy plus kinetic energy.
This small e is a specific internal energy
plus specific kinetic energy, so that is i
plus half u i u i is what this internal energy
is.
But we can also write in terms of other energy
equations. For example, in terms of enthalpy,
this may be useful if you are doing for example,
turbo machinery application and in such cases
the enthalpy difference will give you the
amount of work that is done by the system
on the turbine blades, which is then use for
extraction of power. So, in such cases, for
example, you could be using enthalpy. And
through manipulation of this and other equation
of states and those kinds of things, you could
rewrite this in the form of enthalpy like
this.
And the form is very similar, you have dou
by dou t of dou e here and you have dou by
dou t of rho h the enthalpy. And you have
dou by dou x j of rho e u j this is what we
call as a advection term, so this is associated
with flow. The fact that flow is coming in
and going out of the control volume brings
with it anything that the fluid process including
the x-momentum, y-momentum, z-momentum, internal
energy, enthalpy, entropy all those thing
are concentration, all those things are brought
in. And the net thing arising out of the coming
in and going out of the flow through the control
volume. And you have dou p by dou t plus the
rate of work done by stresses - viscous stresses
and heat conduction the net heat conduction
through the surfaces.
And in the specific case of constant properties,
this reduces to this particular form, and
you can neglect this. So, the solution of
energy equation with appropriate boundary
conditions takes account of convective heat
transfer for the fluid. So, the important
thing is that we do not have to do anything
special to simulate convective heat transfer,
it is already included in the governing equation
for the energy, because of this particular
term here, the advection term here is only
place where velocity is coming, and this will
take care of the convective heat transfer.
And solution of this energy equation along
with the governing other governing equations
is sufficient for us to distinguish between
convective heat transfer and how much of convective
its Reynolds’s number dependents, and all
those kind of a things will be to get. By
direct solution of these equations without
having to worry about Reynolds number effect
and all that. Reynolds’s number is included
in the equations; we do not have to bring
it out especially in the form of correlation.
And what we need to do is that we equations
and we identify the domain, and we specify
the boundary conditions so that is all we
need to do, we do not have to do anything
special for convective heat transfer. So,
the boundary conditions what do we want here
we want boundary conditions can be temperature
specified, wall heat flux specified or both
can be specified.
For example, in the case of convective boundary
conditions, which will see later on, we specify
the ambient temperature and a heat transfer
coefficient for example, by natural circulation
from the wall and that will be something like
a mixed type of boundary condition. Whereas
temperature boundary condition is you specify
the wall temperature at the boundary or you
specify the heat flux, you specify the gradient
of the wall of that temperature at the wall
like this.
And it is important that it is not necessary
to specify the heat transfer coefficient,
it would come out as part of the solution.
Now the heat transfer coefficient that we
talking about is the heat transfer coefficient
at the wall inside the flow domain, if the
heat transfer evacuation external to the flow
domain is by a process, which is determined
by the external conditions. For example, you
have a fuel cell stack and you are cooling
it by having a fan blowing over it, then what
is the boundary condition on the walls of
the fuel cell that depends on what is the
heat transfer coefficient that you are creating
outside the computational domain, outside
the wall, outside the fuel cell which is you
are interest.
And what is the role of the fan, for example,
if you put in high speed, then there is more
heat transfer; and if the fan is in low speed,
then there is less heat transfer; if there
is no fan, then it is natural circulation.
So, the heat transfer external to the flow
domain is something if it is relevant you
have to specify. But heat transfer internal
to the flow domain is something that will
come out of the as part of the solution, so
that that is a very important aspect. We do
not has to specify the inside heat transfer
coefficient, all we need to do is to write
down the energy equation in one of the three
forms in as a internal energy or enthalpy
or in the constant property form like this
we can neglect this.
And each of this will give you solution of
this will give you h as a function of x, y,
z; and this will give as a i as a function
of x, y, z. And if we know the C v or C p,
then from the internal energy or from the
enthalpy you can find the temperature. And
solution of this with specified values of
C p and k will directly give you T as a function
of x, y, z. And from that you can compute
the heat fluxes through the walls; and from
that if you still want you can get some idea
of the heat transfer coefficient, but if you
are directly getting heat flux then heat transfer
coefficient is not necessary to be computed
in the CFD solution, it is a post processing
thing.
If you want find out what is the heat transfer
coefficient, then you can do that. It is same
as when we do the calculation of flow field,
we do not specify any friction factor. For
example, in the first example of flow through
the rectangular duct, and the second example
of flow through the triangle duct, we did
not specify any friction factor and we dint
specify that friction factor is given by 16
by Reynolds’s number where or 64 by Reynolds’s
number nothing was given.
We took the governing equation, we apply the
condition of the no slip and that was all;
and from that for the specified pressure gradient,
we got the flow rate. The normal non CFD rule
will have been that you have the specified
pressure gradient, you bring in a friction
factor correlation in terms of the average
velocity, you plug it into the formula; and
from that you get the average velocity. In
order to get the average velocity, you need
to have friction factor specified, but if
you are doing C F D if you are solving the
original equations accurately at every point
within the flow domain, then it is not necessary
to have any friction factor.
If you still want friction factor, you can
deduce it from impose pressure gradient and
the calculated average velocity. In the same
way here, if we specify the wall boundary
conditions, and if we specify the flow rate
and those kinds of things, then you will be
able to get a temperature distribution and
the corresponding heat flux through the walls.
And from the heat flux through the walls and
wall temperature and the ambient temperature,
you would be able to get heat transfer coefficient
if you wanted. So, heat transfer coefficient
need not be specified here, the convective
heat transfer coefficient, it is already taken
care of and it is no where explicitly calculated
unless you want to have it. So, heat transfer
coefficient is not a concept that in with
CFD.
But we are talking about heat transfer coefficient
inside the flow domain. If heat transfer coefficient
outside the flow domain is necessary as part
of the specification of the boundary condition
that is still needs to be done. We need to
specify the heat transfer coefficient because
outside the flow domain, we do not know what
the temperature variation is, we are not computing
that. If you were to include part of outside
domain also in the computation domain then
you do not have to even specify that, but
that is a different story. So, let us not
go too much into that at this stage.
So, for turbulent flow, we have slightly more
difficulty, we will see this in the last module
of this particular course. And what I would
like to mention is that what we are talking
about is the single phase convective heat
transfer. When you have special phenomena
like boiling and condensation, which you know
change of phases, so you have two different
phases or may be several different phases,
multi compounding mixtures, in such cases
we have to do some special treatment, and
those special effects are not included in
this.
So, single phase convective heat transfer
does not require anything to be specified
here except the appropriate boundary conditions
at the walls. The equations already have inside
them, the information necessary to consider
that convective effect inside the flow domain.
Now, what about conductive heat transfer?
Conduction effects in the fluid are automatically
taken account in the form of Fourier law,
because when we wrote down the heat flux through
the walls through the surfaces of this box
like control volume, what is the heat flux
that heat flux is a for a control volume a
control volume, which is embedded in the entire
fluid continuum.
So, there we are talking about heat flux coming
from the neighbouring control volume in to
this because of temperature differences, and
so that is conduction heat transfer within
the medium. So, the conduction heat transfer
within the medium is already included is our
basic formulation in the form of heat fluxes
here.
So, this is a control volume, it has no walls
it is not these are bounding surfaces and
so this is part of the fluid domain with fluid
walls and you have next cell which is here.
And between these two there is a wall there
is a wall which is made up for the fluid itself;
it is not like a physical wall, which is separating
the two.
So, depending on the temperature here and
temperature here; if the temperature here
is more than the temperature here then heat
would come in through this way and q x would
be negative. If this temperature is greater
than this, if this temperature is less than
this, then heat would be flowing out, in which
case q x at this phase would be positive and
that is already included.
Because in going from here to here, we have
made use of the Fourier’s law of heat conduction
that this q here is minus k dou t by dou x,
so that is included in this. So, both conduction
heat transfer and convective heat transfer
inside the fluid domain are included in the
equation that we are solving. And both of
them require only the appropriate boundary
condition to be specified, so that is what
we are saying here.
Conduction in solids should be taken account;
conduction effects in the fluid are automatically
taken account in the form of Fourier’s law
which is applied to evaluate the conductive
heat transfer within the fluid continuum.
Now if you are talking about the solid for
example, we have a fluid here and then the
solid in between and then the fluid here,
then they can be heat transfer from this hot
fluid into the cold fluid through this intervening
wall, and so our computational domain will
be partly this and partly this separated by
the wall. But heat is transferring from the
bulk of this fluid to the tube wall and from
the tube wall through the other side of the
tube wall into this. So, there you have what
is known as a conjugate heat transfer problem.
So, in such cases the conduction through the
walls needs to be considered by an equation
like this. It is essentially the same as the
equation for the fluids except that this term
is 0, and this term is 0. And s here refers
to an arbitrary source term there may be some
heat generation terms which may be present,
for example, due to a chemical reaction or
due to some friction or due to some nuclear
reaction that may be going on. So, in the
absence of this then t this will give us the
equation for transient heat conduction.
So, conjugate heat transfer involving convection
the fluid and conduction the adjacent wall
is handled through the boundary condition
at the interface, such that minus k gradient
of temperature within the solid, which represents
the heat flux from the solid side of the wall
and that must be equal to k d minus the gradient
on the fluid side of a wall.
And so these two fluxes must be the same at
the interface which is separating the solid
liquid interfaces solid fluid interface. So,
this becomes a boundary condition for the
solids and this becomes the boundary conditions
for the liquids. We do not know exactly what
this wall temperature is, but it can it eventually
be found out when we solve the entire problem
involving convective, conductive heat transfer
within the fluid and conductive heat transfer
within the solid. So, we will come back to
these in a tutorial problem.
Now, of the two modes of the heat transfer
convective heat transfer and conductive heat
transfer nothing more needs to be done from
a CFD point of view. We need to identify the
domain, we need to identify the equation the
extra energy equation, and bring out the essentially
the specific heat and thermal conductivity
of the liquid of the fluid as a special properties
that need to be specified. And we also need
to specify thermal boundary condition at the
surfaces which are bounding this flow domain
computational domain.
When you come to radiative heat transfer,
we have an entirely different based; usually
radiative heat transfer is not taught that
much in our under graduate curriculum. So,
I would like to just put in some simple facts
related to radiative heat transfer for us
to appreciate what is the important about
it. So, this is the dominating mode of heat
transfer when the temperature is high. For
example, in the furnace, where the temperature
can be 1000 degrees, 500 degrees, 1500, 2000
degrees centigrade. And at such high temperatures,
radiative heat transfer is the dominant mode
of heat transfer and unless this is done properly,
we would not get a proper temperature distribution.
So, in such cases radiative heat transfer
become an important issue to be tackled.
Now, thermal radiative energy transfer occurs
within the ultra violet the visible and the
infrared region of the electromagnetic spectrum.
So, we are looking at 0.01 micron to 100 micron
region, and I would like to bring out the
very thin spectrum which is 0.38 to 0.76 micron
within which the visible light comes and then
the infrared region goes from 0.76 to 100
micron.
So, it is a very wide spectrum for the infrared
region and we have fortunate to have eyes
which are sensitive to this very thin band
through which sunlight most of the sunlight
energy is received. So, anyway that is beside
the point. For a body between 500 and 1500
degree centigrade more than 98 percent of
radiant energy falls within the infrared region
of 0.76 to 25 minus micron. So that means,
that within this range 98 percent of the total
radiant energy that is being emitted by body
at temperature which are up to 1500 degree
centigrade or less, then most of the energy
is going to in the infrared region and that
spectrum needs to be resolved.
Solids and liquids emit radiation over the
entire spectrum, mono atomic and diatomic
gases such as O 2, N 2 and argon, radiate
weakly, they do not participant significantly.
Polyatomic gases including carbon dioxide,
steam, ammonia, sulphur dioxide, and hydro
carbons emit and absorb radiation appreciably,
but usually in certain bands of wave length
called spectral windows.
So, why are these gases important? Because
when we are dealing with heat transfer applications
like furnaces, then and if we are looking
at hydrocarbon fuels then the carbon gets
oxide to carbon dioxide, the hydrogen in the
hydro carbon fuel becomes steam. And under
certain conditions, we can have ammonia produced,
and sulphur dioxide may be produced, if sulphur
is present in the fuel. So, these are also
special gases which participate in the radiative
process, whereas oxygen nitrogen and argon
these are essentially transparent, they do
not participate, they let the radiation go
through without interacting with the protons
that are passing through
But and these are also special kind here,
because only in certain bands of wave length
do these gases absorb and radiate and reradiate
and scatter and interact with the radiation.
And whether this is the role of these things
in the role of radiatory transfer significant
a significant depends on the temperature.
And whether there is significant amount of
radiation in which these bands are the spectral
windows of these gases are important. So,
non-luminous gases are the dominant mode of
heat transfer, and combustion products from
combustion production in natural gas-fired
plants. For solid liquid fuels, luminous radiation
is emitted by soot ash and unburnt fuel droplets
and particles.
Again when we are looking at furnaces, we
can have gases fuels, and we can have solid
fuels and liquid fuels. So, the solid fuels
and liquid fuels are typically heavy hydro
carbons, so that means that in the process
of combustion they go through lots of chain
reactions and produce intermediate compounds,
and they may also have more of inert materials
inert solid materials which comes out as in
the form of hash.
And the structure the fuel may be sufficiently
complex that all the fuel is not released
nor all of it is burnt, so you have unburnt
fuel droplet or particles. And these small
sized particulate substances are can have
a significant role in radiative heat transfer,
and they can be like bright sparks, they can
produce luminous radiation, whereas gases
are essentially non-luminous radiation.
Every hot body produces emits radiation according
to the Planck’s law, where the energy see
we have E b is the total energy emission from
point on a black surface to all directions
within the hemisphere about it. And so monochromatic,
so that is the given wavelength emissive power
of a black surface or that of a perfect emitter,
these are all radiation heat transfer language
is given by the Planck’s law where you have
two constants with which huge numbers. But
you have lambdas which are in microns, so
the lambda is very small here. And so integration
over all wavelengths gives the Stefan-Boltzmann
equation. It is the total radiation radiant
energy emitted by a body at a certain temperature
T is given a various as T to the power 4,
so the higher the temperature the more will
be the radiant energy.
It is true of even conduction convective heat
transfer, but there the amount of heat flux
is proportional to the temperature difference.
And here it is two the power of 4, T to the
power of 4, so that means, that this can be
this can increase rapidly. And fortunately
we have the Stefan-Boltzmann constant which
10 to the power minus 4, so this becomes significant
only when temperature is pretty high so at
high temperatures because of the dominant
mode of heat transfer.
So, if you want to predict radiative heat
transfer considers a beam of photons which
is passing through medium here. And this medium
can be a solid medium, it can be gases medium,
or it can be a liquid, or it may be gases
medium with lots of particles ash particles,
soot particles, drops or different sizes and
all of these can interact with the photons.
And they can either absorb them, so that that
particular photon does not come out, and the
energy associated with that photon is taken
away by the absorbing medium, this medium
so that it becomes hotter. Or it can scatter
because this photon is going in straight line
here it can be deflected out in a different
direction, so that this is not accounted for
in this particular beam.
Or the medium can scatter other beams in the
direction of the same beam here, so that you
can have in scattering or out scattering;
while in scattering increases the intensity
of this radiation as it is going, out scattering
will decrease the radiation. And the medium
itself because of its temperature and because
of the Planck’s law and all that it will
be emitting radiation in a certain wavelength
of a certain length as per the Planck’s
law. So, all these things together can be
written in the form of an integral of differential
equation.
So you have differential part is coming from
this gradient here and then the integral part
is coming from the integration of the scattering
kernel. And then you have absorption here.
In scattering, out scattering, and then emission
from the medium, and absorption all these
thing will come out in the form of an integral
of differential equation. So, this equation
is the very different from the equation that
we have considered and we have to solve this
in order to get a proper treatment of radiative
heat transfer, so that there are number of
methods that have been developed and these
are very different from the usual CFD type
of solution methods. Under specialised cases,
these can become like that C F D kind of things,,
but that is very rare.
So, you have very special methods that are
used to solve this equation. And if you want
to do the radiation heat transfer properly
unlike in the case of conductive and convective
heat transfer, we have to solve an extra equation
which is extremely difficult to solve in the
conventional CFD framework. So, you have to
do this and only then you can get proper treatment
of heat transfer in high temperature flow
situations as in the case of furnaces and
boilers and those kinds of things we have
to consider this.
So, in the next class, we will look at mass
transfer. What we have done about conductive
heat transfer, convective heat transfer, all
this is sufficient for us to deal with conventional
applications involving essentially low temperature,
ambient temperature, heat transfer applications.
We will look at how we can deal with mass
transfer and what more is required for the
mass transfer, because that is also an important
aspect when we dealing with industrial applications.
Then we will see what kind of problem formulation
is needed to deal with mass transfer. We have
seen that when we want to deal with heat transfer,
we have to bring in the energy equation, and
we have to bring in the temperature from that.
Similarly, what do we from the mass transfer
that is what we will do in the next class.
