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The fundamentals of transport processes, this is lecture number nine and last time we were
discussing unidirectional flows. Just to remind you of where we started, I first explain to
you why transport processes are important in material transformations both physical
and chemical. It is not just sufficient to put in the raw materials, put in the heat
required, the product need not come out. You need to make sure that the materials get to
the place where the transformation actually occurs and that is where the discussion of
transport phenomena is so important. We looked at dimensional analysis and empirical correlations
based on dimensional analysis and I try to give you some physical inside into what those
dimensionless groups mean. And we talked about diffusion, the molecular
phenomenon of diffusion and how that translates into diffusion coefficients, why the values
of the diffusion coefficients are the numbers that they are in gases and liquids. And then,
we got down to the business of solving actual problems in transport phenomena. The simplest
configuration you can consider is unidirectional transport. That means the transport occurs
only in one direction and that is why the problem is greatly simplified.
And in what is called a Cartesian coordinate system. The three coordinates are x y and
z and unidirectional transport in a Cartesian coordinate system means that it, the transport
is taking place only along one direction and the simplest example in Cartesian coordinates
is the transport between two flat plates and we looked at various instances of heat and
mass transfer between two flat plates both at steady state and at unsteady state.
So in the simplest example of heat transfer; you have two plates one of which is at a higher
temperature and the other of which is at I am sorry this is at a lower temperature and
this is at a higher temperature. Because of this temperature difference energy is transported
from the higher to the lower temperature and that results in a temperature gradient. In
the final steady state the temperature gradient is going to be linear, it is going to be a
linear function of position you’ll have a linear variation in temperature between
the two plates which means of the flux is a constant because flux is proportional to
the derivative of the temperature. However, one could also consider unsteady situations
for example, you have this entire system at one constant temperature T naught and instantaneously
you change the temperature of the bottom surface to T 1 and the temperature is going to change
as a function of time. It is going to look something like this. At very early times and
as time becomes larger and larger you are going to get a flat profile in the long time
limit. Same thing with concentration you can have two plates; one at a lower concentration,
one at a higher concentration and it is going to be a flux from one plate to the other.
At steady state when there is nothing; changes in time that flux has to be a constant and
that concentration has to be linear function of position between the two plates.
We looked at a momentum transport problem. In this case one plate is stationary, it has
no velocity the other plate is moving with a constant velocity. In the final steady state
of course, the velocity profile is linear and the momentum flux or the shear stress
is going to be a constant throughout. But, we also want to look at what happens when
there is a variation in time and that was the purpose why we started looking at what
are called shell balances.
The ideas as follows; we write a balance equation for a small differential volume within the
fluid. We assume that the length the thickness of this volume as well as the area of this
volume is small compared to the macroscopic scale which in this case is the total length
theorem. It has to be small compared to l but, it has to be large enough that the continuum
approximation is valid. In other words it has to be large enough that it contains a
large number of molecules going back to our earlier discussion of the continuum approximation.
If it contains a large number of molecules then you can define continuum fields. In this
case the energy density field. So because of the of the of the of the gradient in the
difference in temperature there’s going to be energy coming in and going out of this
volume. The bottom surface of this volume is hotter than the top surface therefore,
the flux at the bottom will there will be a flux coming in at the bottom and a flux
going out at the top. Energy conservation condition basically set the change in energy
within this time delta T is equal to energy in minus energy out simple. Just a balance
telling you that there can be no creation or distraction of energy. This is valid only
when there is no a source or syncs. If you imagine a chemical reaction which was generating
heat there would be a source. If there is an endothermic reaction which absorbed heat
there would be a sync of energy. If none of these is present you just have a fluid which
neither generates or consumes energy then the balance condition basically says the change
in energy in the time delta T is equal to energy in minus energy out.
And we calculated explicitly the change in energy within a time delta T the energy within
this volume is equal to the specific energy energy per unit volume at this point x y z
times the small volume around it. Therefore, the change in energy was written as e at x
y Z T plus delta T e at T x y Z T plus delta T minus e at x y Z T divided by times the
volume itself and that we wrote in terms of the specific heat and the temperature here.
The fluxes in and out is equal to the heat flux. The flux now is in the Z direction because
there is a variation temperature only in the Z direction. So, the energy in this q Z times
delta x times delta y that is the, q Z is the flux energy per unit area per unit time.
q Z times delta x delta y is the energy in per unit time which is the flux times the
area and then multiply that by delta T to find out how much is coming in within that
time interval delta T put all of these together.
And we got a balance equation which was basically of this form rho c p into d T by d T is equal
to minus partial q Z by partial z. So this basically tells you that the rate of change
of specific energy within that volume is equal to the rate at which energy comes in minus
the rate at which it is going out. The difference in fluxes across the two surfaces the bounding
surfaces.
And then we need an equation for the flux. Fourier’s law of heat condition q Z is equal
to minus k partial T by partial Z. Note the partials here. These are important. It implies
that you’re taking the variation in temperature with Z keeping all other coordinates the same.
That is you are varying Z a little bit and finding out the difference in temperature
keeping x y and T a constant. And from that we got this diffusion equation for the temperature
where alpha is the thermal diffusivity.
Did exactly the same thing for mass. Instead of T we have c the concentration, instead
of alpha we have d the diffusion coefficient and you get the exact same equation.
The exact same equation except that concentration is substituted for temperature; mass diffusion
coefficient is substituted for the thermal diffusion coefficient. In all of these we
assume that the diffusion coefficients both from mass and the thermal diffusion coefficients
are not functions of position. If they are functions of position it gets a little more
complicated. We will assume that the fluid properties are uniform so the thermal conductivity,
the mass diffusivity does not depend upon position.
The form of the equation for momentum diffusion was slightly different. It came out of Newton’s
law third law, Newton’s laws of motion rate of change of momentum. In a differential volume
is equal to the some of the applied forces and there I had defined for you in some detail
what is the stress tau x z?
It is the force per unit area in the x direction acting at the surface with outward note outward.
You are at normal in the Z direction and I also told you that if the outward unit normal
is in the minus Z direction the force will be minus tau x Z because if you interchange
if you if you if you reverse the direction of the unit normal the direction of the force
also changes according to Newton’s third law.
And then we have, we basically use the balance condition that the rate of change of momentum
is equal to sum of the applied forces to obtain an equation for momentum transfer which once
again looks identical to the equations for heat and mass transfer. Except instead of
concentration you have the velocity u x. Instead of mass diffusion coefficient you have the
kinematic viscosity or momentum diffusivity nu; ratio of viscosity and density.
And we talked a little bit about scaling. I said that rather than work with these un
scale coordinates alpha I am sorry T u x and c right I could define scaled coordinates.
In this case T star is equal to T minus T naught by T 1 minus T naught. T star is equal
to 0 on the top surface and T star is equal to 1 on the bottom surface. If I define Z
star is equal to Z by l then Z star is equal to 1 on the top surface and Z star is equal
to 0 on the bottom surface. Define this way; in all cases the concentration the scaled
temperature varies between 0 and 1. The scaled concentration varies between 0 and 1 the scaled
velocity varies between 0 and 1. So the equations are the same except that
one has to be put in the appropriate diffusivity the lengths all scale vary between 0 and 1.
The the the quantity is being transported. Once again vary between 0 and 1. So, all of
them have a common solution. This is the solution at steady state. The linear profile for the
temperature concentration and velocity fluids. So, that is where we stopped off the last
lecture. Now, we come to looking at the simplest unsteady state problem. That is the problem
that I would just explained to you a little earlier if initially the temperature is equal
to 0 everywhere and at some time T is equal to 0. The temperature at the bottom surface
is increased to one then, what happens? How can you can we measure the temperature field
in this unsteady situation? That is the topic for today’s lecture.
So let us go ahead with analyzing that problem. So, back to our problem two plates and I will
just work in scaled coordinate Z star is this axis. I can scale by the thickness whatever
the thickness is, I can scale it by the thickness to get this boundary at Z star is equal to
0 and this boundary at Z star equals 1. So, the scale distance varies between 0 and 1.
The boundary conditions in this case are as follows for T less than 0 right, I will assume that the
concentration the I am sorry the temperature T star is equal to 0 everywhere. So, it is
equal to 0 here and it is equal to 0 on this surface as well and at T is equal to at T
is greater than or equal to 0 T star is equal to 1 at Z star is equal to 0. So, at the initial
time instantaneously so initially if I plotted the temperature; the temperature was basically
equal to 0 everywhere. Scale temperature I showed you in the last class how to scale
it. So, this is the temperature T star is equal to 0 at the initial time. I instantaneously
increase the temperature here to T star is equal to 1. So, initially the fluid has not
felt the heat coming out of this plate. So, it is just at 0 temperature. Gradually there
is going to be a diffusion of heat from the bottom because I have raised the plate of
the raised the temperature of the bottom plate. So, initially I am going to get a temperature
profile that looks like this. That temperature profile is slowly expanded as a heat goes
further and further up until in the longtime limit. I am going to get back a linear temperature
profile. Let me just put that little more carefully. The longtime limit I am going to
get this linear temperature profile and I want to find out how the temperature evolves
from this initial state to the final linear state before we tackle the problem of the
temperature evolution from the initial to the final steady state. Let us look at the
very early instants of time when the temperature disturbance due to the bottom plate has not
yet gone far into the fluid. So, if the temperature disturbance at the bottom plate generated
due to instantaneously increase in the temperature of the bottom plate if it has not gone very
very far up then this disturbance is going to be restricted to what is called a penetration
depth. It is going to be restricted to a penetration depth which is small compared to the total
height the total height was Z is equal to 0 to Z is equal to l. If the penetration depth
is small compared to that; that means that the as as as far as the temperature field
is concerned, the top surface is very far away. So, I do not have to know the exact
location of the top surface. I just have to know that as I go further and further up.
The temperature reaches 0 as the distance becomes very large ok.
So my problem rather than T star is equal to 1 at Z is equal to 0 and T star is equal
to 0. Everywhere if the penetration depth is small compared to the total height then
I can say that T star is equal to 0 as Z goes to infinity. When I go further and further
up the temperature approaches in 0. In case the temperature in this case is approaching
0; in this case it is approaching 0 as I go further and further up. So, in that sense
I am effectively solving a problem in an infinite domain. The problem reduces to a problem in
an infinite domain because the penetration depth to which the the the temperature disturbance
due to the bottom surface has been felt is very small compared to the total height.
So, this is the temperature boundary conditions that I will use and the differential equation
for the temperature field that we had got was d T by d T is equal to alpha d square
T by d Z square. So, this was the differential equation when we initially scaled variables.
We scaled it by the total height of this. The distance between the two plates H. We
will scale it by the total distance between the two plates. However, now I am considering
a situation where the distance to the top plate is large compared to the penetration
depth within the flow. So, I am effectively enforcing boundary conditions in the limit
as Z goes to infinity. If the penetration depth is small compared to the height it does
not matter what the height is. It can be H, it can be 2 H as long as a penetration depth
is small. The only requirement for the temperature field is that the scale temperature goes to
0 as H goes to as as Z goes to infinity. So previously, because I had a finite depth
I scaled all length scales by H. In the present case H is no longer a parameter in the problem.
So, how do I scale the distance Z? Let us go back and look at all the relevant variables
in the problem. One relevant variable of course, is the temperature T star but, that is already
dimensionless. T star at scaled as T minus T naught by T 1 minus T naught varies between
0 and 1. No units; it is dimensionless so that cannot be a dimension. That is that is
important in this case. I have three others; two are the coordinates, one is the z coordinate,
the other is time and the only other thing that I have in the problem is this diffusivity
alpha. So there are there is one dimensionless variable
T star, two coordinates Z and T and 1 dimensional parameter, the thermal diffusivity alpha.
So, I do not have any length scale to scale Z by because initially at Z that the distance
between the plates was H. But, then if the the the temperature field hasn’t penetrated
that far right; it does not matter what H is. The temperature field near the bottom
surface is not going to depend upon what the total height is so that is the issue here.
Now, the answer to this problem is as follows. I have three variables T, Z, alpha which have
dimensions and there are only two dimensions that they can depend upon. They do not depend
upon mass here, they depend only upon length and time. Therefore, from these three coordinates,
I can get only one dimensionless group. I told you last class that we tried to we tried
to a scale or variables. When we try to solve this problem so if I if I did the problem
simplistically what I would do is to scale Z by a height and T by a time. In this case
I do not have a height because the penetration depth is small so the the boundary conditions
applied at Z going to infinity. I do not have a scale for time. Either out of these three;
I can get only one dimensionless group and that dimensionless group is a similarity variable.
That dimensionless group is a similarity variable Z by root alpha t.
So just from dimensional analysis; if the penetration depth is small so that I am applying
a boundary condition at infinity the progression of the temperature field with Z and with time
are linked and they are they are linked in such a way that the progression depends only
upon this similarity variable or similarity coordinate and that effectively reduces the
problem from two independent variables Z and T to just one independent variables which
is a dimensional necessity. I am sorry which is a dimensional necessity because there are
no other dimensionless groups that I can form in the problem.
So, if the temperature field depends only upon this a dimensionless similarity variable
then I should be able to express my entire equation in terms of that dimensionless similarity
variable alone. So, what I need to do is to express Z the derivative with respect to Z.
In terms of the derivative with respect to psi the derivative with respect to T also
can be expressed in terms of the derivative with respect to psi. Put that into this equation
and if my premise is correct that it depends only through this variable psi and not individually
through T Z and alpha. The final equation that I get should not contain any of these
parameters T Z or alpha. It should be only a function of psi alone so that is that is
the basic idea ok.
So, let us proceed with the solution. So, I have my equation partial T star with partial
T is equal to alpha partial square T star by partial Z square and I have defined my
similarity variable psi is equal to Z by square root of alpha T. This is dimensionless because
Z has dimensions of length alpha has a length square per unit time. Therefore, alpha T has
dimensions of length square and therefore, Z by root alpha T is dimensionless. So, the
derivative with respect to time partial T star by partial T is equal to my partial psi
chain rule for differentiation partial psi by partial T is equal to minus Z by 2 root
alpha 2 power three half’s times partial T star by partial psi. Now, Z by root alpha
T is psi itself Z by root alpha T is psi itself. Therefore, I can also write this as minus
psi by 2 T partial T star by partial psi using chain rule for differentiation.
On the right hand side partial T by partial Z; first take the derivative respect to Z
and then with respect to psi by chain rule this is 1 over root alpha T partial T by partial
psi. In the second derivative, partial by partial Z of partial T by partial Z is equal
to partial psi by partial Z partial by partial psi of and this is equal to 1 by alpha T partial
square T by partial psi square. Now, I have to substitute this into the original diffusion
equation therefore, I will get partial, I will rub it out, I will get minus psi by 2
T partial, T by partial psi. This is equal to alpha by alpha T partial square T by partial
psi square and I can cancel carefully and you can cancel out alpha there and T here
to get an equation of the form minus psi by 2 partial T by partial psi is equal to partial
square T by partial psi square good. So, just to recall I had previously said that since
by dimensional analysis there is only one dimensionless group which involves both the
Z coordinate and time I should be able to define this entire, I should be able to rewrite
this entire equation in terms of this dimensionless group alone.
Once I have written it that way the equation should not contain individually Z T and alpha;
it should only contain this dimensionless variable psi and I do find indeed that the
final equation that I get thus contains only this dimensionless variable psi.
But, however if I am to solve this equation; I need boundary conditions. What are the boundary
conditions that we had at Z is equal to 0 T star is equal to 1? So, at the bottom plate
the temperature T star is equal to 1 at Z is equal to 0. So, that Z is equal to 0 is
equivalent to psi equal to 0 because psi is equal to Z by square root of alpha T. The
other boundary condition was that in the limit as Z became large Z going to infinity T star
was equal to 0. The limit as Z goes to infinity T star was equal to 0 Z going to infinity
is equivalent to psi going to infinity psi going to infinity because Z psi is equal to
Z by square root of alpha T. That was a third boundary condition rather an initial condition
and that was that at T is equal to 0 at the initial time when I just switched on this
temperature field at the bottom. Everywhere else within the flow within the fluid the
temperature was equal to 0 at the initial time T is equal to 0 when I just switched
on the heating from the bottom the temperature was equal to 0 everywhere else apart from
that bottom plate. So, that would imply that at T is equal to 0 for Z greater than 0 T
star is equal to 0 right at time T is equal to 0 everywhere in the fluid apart from that
bottom plate apart from Z. Z is equal to 0 the temperature is equal to 0 T is equal to
0 psi is equal to Z by square root of alpha T. So, this implies that this has to be applied
at psi going to infinity and you can see that when expressed in terms of the psi coordinate;
these two boundary conditions are exactly the same. I had started off with a second
order differential equation in the Z coordinate, the first order differential equation in time
that required two boundary conditions in the special coordinates. One initial condition
in time and then I reduced the variables using the similarity transform to an equation just
in terms of one variable psi. This equation was only in terms of the variable psi and
because it is a second order equation in psi. You can have only two boundary conditions
for this equation and therefore, what we find is that one of the boundary conditions and
one initial condition of the original problem turns out to be exactly the same when re expressed
in terms of psi. So, that we get a consistent solution.
The other thing that I have done here is to convert a partial differential equation with
variations in time as well as Z into an ordinary differential equation and that is the big
advantage of this method. The similarity solution was defined so that the partial differential
equation which I originally had is in transform coordinates and ordinary differential equation
and ordinary differential equation is simpler to solve. It is a linear ordinary differential
equation and so I can solve it to get a solution for the temperature profile.
So my equation is minus psi by 2 d d T star by d psi is equal to T square T star by d
psi square. So, I can solve this. Now in case so if I define u is equal to d T star by d
psi then I will get minus psi by 2 u is equal to d u by d psi and u is equal to a constant
e power minus psi square by four and this is equal to d T by d psi. Therefore, T is
equal to c integral 0 to psi d psi prime e power minus psi prime square by four plus
some other constant d. And I have boundary conditions T star is equal
to 0 as psi goes to infinity T star is equal to 1 at psi is equal to 0. Using this, I can
easily determine the two constants. Using these two boundary conditions I can determine
the two constants in equation.
So this will give me T star is equal to 1 minus integral 0 to psi d psi prime e power
minus psi prime squared by four divided by integral 0 to infinity prime square by four.
That is what I would get after I enforce both boundary conditions and I put in the solution the the the similarity
transform. Once again that similarity transform you will note comes in only in the limits
of this integration. Therefore, in terms of the original variables I can write T star
is equal to 1 minus integral 0 to Z by root alpha T d psi prime e power minus psi prime
square by four by integral 0 to infinity d psi prime. So, that is the final solution
for the temperature field. The temperature field depends on position and time only through
this scaled coordinate. And now we are in a position to answer the
question that we had originally asked what is the penetration depth? Clearly the temperature
as Z by root alpha T becomes larger and larger in the limit as it becomes large it goes to
infinity. The temperature will go to 0, the temperature deviates significantly from 0
only when Z by root alpha T is approximately 1, is of order 1. If it becomes large there
is virtually no deviation from 0 therefore, the penetration depth in this case is proportional to square root
of alpha times T it goes as square root of alpha times T.
So, the penetration depth is increasing as square root of T as time increases. The next
question what did we mean initially when we said that the time is small enough that the
penetration depth is small compared to the total height penetration depth is small compared
to the total height. Only when square root of alpha T is small compared to the height
H or when T is small compared to H square by alpha. So, this is the time period over
which you can use a similarity solution when the transport problem looks like the transport
into an infinite fluid because the height is large compared to the penetration depth.
So this gives me the temperature profile at every instant in time, a second relevant question
that one can ask is for example, what is the heat flux at the bottom surface? The heat
flux
q is equal to minus k times d T by d Z k is a thermal conductivity expressed in terms
of the scale temperature. This will be equal to minus k into T 1 minus T naught into partial
T star by partial Z because T star is T minus T naught by T 1 minus T naught. Therefore,
d T star by d Z is d T d Z divided by T 1 minus T 1 good.
Now, this we have to express in terms of the scale variable psi. So, this when expressed
in terms of the scaled variable psi becomes T naught into d Z d d psi by d Z times d T
star by d psi and d psi by d Z is 1 over square root of alpha T this equals to minus k into
T 1 minus T naught by square root of alpha T into d T star by d psi the flux at the bottom
heat flux at Z is equal to 0 Z is equal to 0 also corresponds to psi is equal to 0 because
psi is equal to Z by square root of alpha T so q Z at Z is equal to 0 is equal to minus
k into T 1 minus T naught by root alpha T into d T by d psi at psi is equal to 0.
And now, you can use this solution for T in order to find out what is d T by d psi. You
can clearly hear this constant as 1 over integral 0, 2 infinity of d psi prime. So, I use this
solution the derivative with respect to psi will just remove this integral sign on the
second term and substitute psi is equal to 1 here. So, this is equal to minus k into
T 1 minus T naught by square root of alpha T into into minus 1 by integral 0 to infinity
d psi prime e power minus psi prime so squared by four. So, this basically is equal to k
into T naught minus T infinity by root alpha T integral 0 to infinity d psi prime e power
minus psi prime square by four. So, this term here this is just a constant it is a definite
integral from 0 to infinity. So, I am getting the heat flux decreasing with time as 1 over
square root of T the penetration depth increases the heat flux is the gradient of the temperature.
It will go approximately as the temperature divided by the penetration depth as a penetration
depth increases the heat flux decreases as 1 over square root of T in this particular
problem. So, to recall we use similarity transform
because there were no length or time scales in the problem with which we would scale length
and time by and because of that we manage to get a solution which was only in terms
of a similarity variable. Now, this similarity transform can also be applied for mass and
momentum transport equations in problems. The only thing is that I would substitute
the mass diffusivity for the thermal diffusivity. Here in my similarity variable I would substitute
the momentum diffusivity kinematic viscosity nu for the thermal diffusivity here in this
problem.
So for example, if I considered mass diffusion problem where there was diffusion from a surface
c star is equal to 0 here and c star is equal to 1 here and initially c star was 0 everywhere
and then I considered the problem of early times you start off initially with this profile
then you you had a profile like this where the penetration depth was small compared to the total height right
is exactly the same equation and I can write down the solution c star is equal to 1 minus
integral 0 to Z by root d T d psi prime e power minus psi prime square by 4 by integral
0 to infinity. So, in this case the concentration was 0 everywhere
at the initial time T is equal to 0 I said the concentration of the bottom plate alone
equal to 1 and then watch the diffusion from there equivalent momentum transport problem.
So, I have u x is equal to 0 everywhere at time T equal to 0; I set this velocity into
motion with a velocity u in case that u x star which is a scale velocity is equal to
1. In the final state of course, I will have a linear velocity profile but, in the very
initial stages the momentum transfer will be limited to a thin region near the base
where the penetration depth is small compared to the total height H. So in this case, I can also write down quite
easily what the solution for the velocity is going to be equal to square root of Z by
root nu T where nu is the kinematic viscosity. So, you get exactly the same solutions. These
are semi infinite domains where you have one surface an an infinite fluid on top and either
the concentration temperature of the velocity of the bottom surface instantaneously changed
and then you find the diffusion going upwards. So, we do that by this similarity transform.
In this case it was a dimensional necessity the the fact that there is only one dimensionless
group was a dimensional dimensional necessity. However, this similarity transform is more
general than that we will use it later on when we do boundary layer theory. We will
repeatedly use similarity transforms to reduce the equation from a partial deferential equation
to an ordinary differential equation.
And I can show you one example here where such a thing is done. We will consider here
the simplest example where we can use the similarity transform not based upon dimensional
analysis but, rather based upon physical understanding and that is the mass transfer to a falling
film of fluid. Let us say that I have a film of fluid falling
downwards along the wall of of of some contactor and there is gas here, there is gas at the
surface. So, at the entrance so the the the there is there is an entrance of towards the
fluid then it follows falls as a thin film along the wall and it comes into contact with
gas along the wall in the idea is to dissolve this gas into this film of fluid. Now, the
film of fluid in general will have some velocity profile like this. We will assume that this
width of this film is H and the total length one that it travels is now l and there is
this velocity profile very close to the surface. Now, clearly as the gas in the film at the
entrance the concentration of the the the the solute that is being absorbed absorbed
from the gas c will be equal to 0. At the entrance as the solution proceeds the concentration
increases as you go downwards at the interface itself the concentration is a constant value.
It is equal to the equilibrium concentration. The solubility c s in that is in equilibrium
with the gas at its particular temperature pressure and concentration so we put a coordinate
system here as usual. I will take the depth here as Z depth inverts into the film and
this stream wise flow direction as x. And now, I want to find out what is the concentration
in the liquid as function of time my final objective is to get a Nusselt number correlation
for this; a correlation similar to the correlations that we had discussed when we did dimensional
analysis. Can we get a correlation for this as a function of l, the total length H, the
fluid velocity and the diffusivities? So, that will be the final objective this is now
not an unsteady problem. It is a steady state problem. However the concentration is now
changing as a function of x. So, transport in this case is no longer unidirectional.
There is a variation of concentration in the Z coordinate. There is also a variation of
concentration in the x coordinate. The reason that I am doing it at this stage where we
are discussing unidirectional flows will become apparent a little later on. Turns out that
the solution for this configuration is exactly the same that you get for an unsteady flow
into an infinite fluid. Now, as the gas comes in contact you would
expect that initially the concentration is at is is is 0. In the fluid as it comes into
contact there will be a thin layer of fluid at the surface where the effect of this the
dissolution of the gas at the interface is felt this is the equivalent of the penetration
depth. In our previous case in that case the penetration depth increases with time as as
time progresses the the the solute at the bottom dissolves further and further into
the liquid. In this case it is as a function of the x coordinate as as time increases the
the the gas at the interface dissolves further and further into the liquid that is trying
to dissolve the the gas. Now, so I have two coordinates here z and
x and I have a velocity field I am going to make an assumption and that is that this velocity
is a constant as far as the flow is concerned. I am going to make an assumption that u is
equal to constant. This is an assumption this is valid if and only if the depth to which
the concentration has penetrated is small compared to the total height H. in our In
other words, I require that the change in velocity between these two positions has got
to be small compared to the magnitude of the velocity itself. So, if I expand out this
region alone I have this interface I have this penetration depth here and then I have
a velocity profile along this. I have a velocity profile that goes like that my constant velocity approximation is valid
only if the change in velocity in this region is small compared to the total velocity itself.
If the change in velocity is small; I can assume that the velocity is approximately
a constant over the thickness over which there is penetration of the gas similar to our assumption
in the previous case that the penetration depth is small compared to the total height
I have to come back and justify this ok. So, the balance equations are the same as
before except that now we have the system which is at steady state nothing is changing
in time but, there is a change in the x direction. So, I will take a differential volume this
is delta Z perpendicular to the film delta x is along the flow direction. A second assumption
that we will make, we will neglect diffusion along the x direction there is flow along
the direction but, we will assume that there is no diffusion. That is assumption number
two and we will come back and look at under what conditions that assumption is valid for
the moment. We will assume that there is only diffusion only in the Z direction and there
is convective transport because the fluid is flowing downwards there’s going to be
convection in the x direction and a balance between this convection and diffusion is what
is going to give me my differential equation by the shell balance procedure.
So, this is the problem I will do the shell balance in the next lecture and then solve
it and get a solution for this and hopefully from that we will get our first correlation
between the nusselt number and all of the other parameters in the problem. So, we will
see that in the next class. I will briefly review before we leave what we did in the
present class. The solution procedure that we used was what is called the similarity
solution. That similarity solution in this case it works only when there are no length
or times scales in the problem. I had an unsteady problem and Z is equal to 0. I had a boundary
condition. Now, I had another boundary condition as Z goes to infinity and an initial condition
at T equal to 0, there were no length or time scales in this problem.
And therefore, when I post my differential equation I had a deficit of dimensions to
scale my variables with there’s no length scale Z going to infinity, no length scale
in the problem nothing to scale Z with there was no inherent time scale in the problem.
The only thing that I had in the problem was the diffusivity alpha and from the diffusivity
alpha and the Z and T. I could get only one dimensionless group. So, this implied that
just based upon dimensional analysis there is only one independent dimensionless variable
which involves both Z and T and that was related in and that was related as follows.
If this is true then my entire differential equation I should be able to express in terms
of psi alone. The differential equation cannot depend independently on alpha and Z and T.
It should depend upon psi alone and that is what I got. I got a final equation here which
depends only upon psi alone and not individually on Z and T not sufficient for the equations
to be reduced to just an equation for psi. If you would be able to reduce the boundary
conditions as well and the original problem had two boundary conditions; one initial condition,
the problem in terms of the similarity variable should have only two conditions because it
is a second order differential equation in psi alone. And I showed you that one boundary
condition and one initial condition when expressed in terms of psi reduce to the same condition.
Once that was done then I had a way to solve the problem and from that we found that the
penetration depth is increasing as square root of alpha times T and obviously if you
had a finite channel the analysis that we did will be valid only when T is T is small
compared to H square by alpha. At these very initial times this analysis will be valid
when T becomes comparable to H square by alpha then the influence of the of the boundary
condition of the top plate is felt and this analysis is no longer valid and I showed you
what the solution was and the heat flux decreases at T power minus half and I told you that
you can get exactly the same solutions for heat and mass for mass and momentum transfer
I have set up this problem for you. We will continue this in the next class and try to
get a correlation for the nusselt number. We will see you in the next lecture.