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So, welcome to lecture number Thirty Six. So, far discussion of fundamentals of transport
processes. The last few lectures, we first derived the convection diffusion equation,
in different coordinate systems. That equation had the general form dc by dt plus del dot
uc is equal to D del square c plus any sources or sinks that are present, within the flow.
And I showed you that, if you scale the velocities is equal to u by capital U and all distance
x by, some characteristic distance L, the radius of a particle if a particle is inverse
to in the flow, or the pipe diameter, if the flow through the pipe, the time scale becomes
is t equal to t D by L square. In that case, you get a dimensionless equation of the form
Pe into dc star by dt plus divergence of uc is equal to D del square c plus the source.
So, this is what you get and there is the peclet number sitting out in front here, which
is basically, Pe is equal to UL by D. So for, we have been considering the limit of low
peclet number. We have been considering, implies that del square, the equation reduces to D
del square c plus source is equal to 0. So, we look that some general strategies to solve
this problem. D del square c plus the source or sinks within the flows is equal to 0. Basically,
it is general strategies for solving del square c is equal to 0. As the Laplace in of the
concentration field is equal to 0. We looked at different ways of solving this separation
of variables as well as solution strategies based upon considering a point source within
the flow. And then considering the distributed source as the summation of my small point
sources. Here, we are going to start looking at the opposite limit. The peclet number is
large compared to 1. Simplistically, you would think that we can just go head and solve it
by using the equation dc by dt plus divergence of uc is equal to 0. However, that is not
so and I will try to explain to you why that is not.
So, by considering first the simplest case, which is the flow past a flat plate. We consider
solution of the conduction equation in the flow past a flat plate. So, I have a flat
plate here. And T is equal to 0. And at particular location the temperature is instantaneously
increased to a high value. So, T is equal to T1. At this particular location, you starting
to heat the fluid. And the flow near this fluid is a linear velocity profile. The flow
in near the surfaces linear velocity profile. And the temperature of the fluid that is entering
is basically, at the same temperature as this unheated section, T is equal to T naught.
Consider for example, a heat exchanger problem for have a heat coming in at the inlet. I
have the cold fluid coming at the inlet and at one particular location, it comes into
contact with the heated fluid outside. Because, the wall at that particular location is heated.
So, I have cold fluid which is initially coming in at temperature T naught. At one particular
location, I start heating, I set the temperature is equal T1 at the wall of the tube itself.
Rather than considering a cylindrical surfaces, I will just consider a flat plate.
So, Let us first write down our coordinate system. I will use x as the coordinate along
the flow, y as the coordinate perpendicular to the flow. The velocity field, the velocities
only in the x direction, parallel to the flat plate. So, velocity u x is linear function
of y. It increases linear as y increases. I write it as gamma dot times y, where gamma
dot is equal to the strain rate, gamma dot is equal to strain rate. And let us consider
at that, this process is happening at steady state. So that, there is no time derivative.
So, my heat conduction equation just becomes, partial del dot uc is equal to alpha del square, alpha
del square T. This is the equation that I am going to try to solve within this domain.
In the limit of high peclet number. Now, what does high peclet number mean in this case.
So, Let us consider that the heated section has a length L. So, the length L is the heated
section, the length of the heated section. What does high peclet number mean? I have
a thermal diffusion coefficient alpha, dimensions Length square by unit time.
I have characteristic length which is the length of the heated section L, characteristic
length in the problem. And there is no velocity scale but there is a scale for the strain
rate, u x is equal to gamma times y. So, the strain rate gamma dot has dimensions of inverse
time because velocity is length by unit time is equal to gamma dot times the length. Therefore,
the stain rate has dimensions of inverse time. That means, that I can create a dimensionless
number, which is the ratio of convection and diffusion has peclet number is equal to gamma
dot L square by alpha. So, this gives me the ratio of convection and diffusion, in this
particular case. This gives me the ratio of convection diffusion in this particular case.
Further, the velocity is not varying as a function of strain vice direction. So, I have
u x is independent of y. Sorry, independent of x and u y is equal to 0. There is no flow
perpendicular to the plate. So, u x is independent of x and u y, there is no flow perpendicular
to the plate.
So, with this my conservation equation would become u x partial T by partial x. u x is
independent of x. So, I can take it out of the differentiation, plus u y times partial
T by partial y. But u y is equal to 0 is equal to alpha into partial square T by partial
x square plus partial square T by partial y square. So, this is the convection diffusion
equation for the steady case, where there is velocity only in the x direction. So, now,
I scale U x is equal to gamma dot times y, partial T by partial x is equal to alpha,
partial square T by partial x square plus and I have a length scale. So, I can scale
on lengths, scale x star is equal to x by L and y star is equal to y by L. Once, you
do that you will get gamma dot L square by alpha, y star dT by dx star is equal to d
square T by dx star square plus d square T by dy star square. That is the differential
equation in scaled form. Further, I should the temperature as well. I can define a scale
temperature T star is equal to T minus T naught by T 1 minus T naught. In each case T star
is equal to 1 here and T star is equal to 0. So, in terms of the scaled temperature,
since the equation is linear in the anyway, I write in terms of the scaled temperature.
I will just get Pe y star dT star by dx star is equal to d square T by dx square plus.
So, this is the equation that I get with boundary conditions, T star is equal to 1 at y is equal
to 0 for x greater than 0. So, for x greater than 0, I have the heated section of the plate
where T star is equal to 1. So, T star is equal to 1 on this heated section of the plate
in the, at y is equal to 0, for x greater than 0 of course. As I go far from the plate
the temperature should equal to this free steam temperature. If I go very far from the
plate as y goes to infinity, I should recover T star is equal to 0, because the effect of
heating from the plate has not reached that far. As y goes to infinity, T star has to
decreased to 0. And T star is equal to 0, as y star goes to infinity. In addition, fluid
that is incident on the plate for x is less than 0, it has not been heated yet. Fluid
that is incident on the plate for the heated section for x less than 0, it is not yet been
heated. Therefore, for x less than 0 for any value of y greater than 0, the temperature
should be equal to 0. There is the initial condition for the fluid, that is incident
on this heated section. Therefore, I require that T star is equal
to 0 at the location x is equal to 0, because it not touch the heated section of the plate,
for y greater than 0. y is equal to 0 is heated but anywhere above is not yet heated. So,
that the temperature should be 0 for all locations y greater than 0. So, we have to solve this
equation subject to these boundary conditions. In the limit, where the peclet number is large
so that, we would expect convection to be dominant in comparison to diffusion.
So, simplistically, we just neglect diffusion. And we will end up with the equation of the
form dT star by dx star is equal to 0. Because, if I neglect diffusion then I just get Pe
y star times dT by dx is equal to 0. And if y is nonzero, the peclet number non zero,
then dT by dx has to be equal to 0. That means, the temperature variant in the x direction.
The temperature is independent of x. The temperature is independent of x. It depends only upon
the y coordinate. However, that means, the dT star by dx star is equal to 0. This condition,
tells us that T is equal to 0 for all y, at x is equal to 0. Basically, what dT dx is
equal to 0 tells you, there is no variation along the flow stream lines of the temperature.
However, at the inlet the temperature was equal to 0 independent of position. That means,
that the only solution is
T star is equal to 0 everywhere. Since, T is independent of x, at the inlet T was equal
to 0 everywhere. Therefore, the value has to be constant at all values of x. At the
inlet it was equal to 0. Therefore, at all values of x constant as we go downstream,
the temperature has to be identically equal to 0. So, that is the solution that you get
in the high peclet number limit. There is no diffusion. So, because of that there is
no transport across stream lines. No transport across stream lines and because of that, whatever
the temperature was there at the inlet continuous to be the same temperature everywhere, within
the flow at the inlet the temperature was 0. So, everywhere within the flow the temperature
continuous to be 0. Clearly, this solution in not compatible with this boundary condition,
that in the heated section T is equal to 1. Clearly, it is not compatible with this boundary
condition. What this solution is saying, if we neglect diffusion all together the temperature
continuous to be a constant and there is no effect of heating, in that heated section.
So, it is incompitable with one of the boundary conditions. The mathematical reason for this
incompatibility is quite obvious. If I neglect the diffusion terms all together, the diffusion
terms had the second derivative with respect to the y coordinate, the diffusion term at
the second derivative with respect to the y coordinate. That second derivation because
it was second order differential equation in y, I had 2 boundary conditions in y. When
I neglected that diffusion terms, the second order term in y, I converted this from a second
order differential equation to an ordinary equation in y.
So, the ordinary one cannot satisfy any boundary conditions. That is the reason, that we are
not able to get, satisfy the boundary conditions at the surface of the plate. Because, when
we neglect the diffusion terms, we neglect to the highest derivatives. And once I neglect
the highest derivative, it can no longer satisfy all the quite boundary conditions. So, that
is the mathematical reason, why we are not able to satisfy the boundary conditions. The
physical reason is, as I have been telling you numerous times during the course of these
lectures, convections transports mass, momentum and energy only along the flow direction.
Transport across stream lines can takes place due to diffusion. Because convection, convective
transport takes place only along the flow direction. Diffusion has to take place along
perpendicular of flow direction. When I neglect diffusion in comparison to convection, there
is no transport across stream lines. However, if you look at the mean velocity itself at
a bounding surface the mean velocity has to be 0, from the no slip condition and from
the no penetration. There can be no flow perpendicular to the surfaces. At a surface because at the
surface itself, the velocity of the surface has to be equal to the velocity of the fluid.
There is no relative velocity between the fluid and the surfaces at the surface itself.
So, there is no relative motion between the fluid and the surfaces and because and of
that can be no net transport perpendicular to the surface due to convection. Any transport
perpendicular to the surfaces has to ultimately, takes place only due to diffusion. So, because
of this physical reason, because we are neglected diffusion perpendicular to the surface, there
is no transport of heat. And consequently, there is no heating up, change in temperature
due to the heated section, due to when we neglect the diffusion terms. However, diffusion
is to present. It is a molecular phenomenon and it is going to exist. It may be small
but it is still going to exist. So, how does one bring into this problem? And the key basically
lies here. Note, that this condition says that T star is equal to 0 at x star is equal
to 0 for y greater than 0. Note this conditions, it says for y greater than 0. Because, at
the surface itself there is 0 velocity, the velocity is equal to 0 exactly at the surfaces,
there is transport along the surfaces. So, for y greater than 0, T has to be equal to
0. These are the boundary conditions that says
that T star is equal to 1 exactly at y is equal to 0. So, if I plotted the temperature
field due to this, if I plot it as the function of y, if I plotted the temperature at the
surfaces itself the temperature has to be 1, but everywhere above it has to be equal
to 0. That is what the diffusion equation, the convection diffusion equation is saying
without the diffusion terms. However, a profile like this is step profile. If we have a step
profile, the derivative of a step function is a delta function which goes to infinity.
So, if I have a step profile the derivatives are actually very large. So, even though the
peclet number may be a large. And even though, I have an equation of that kind, gamma dot
y dt by dx is equal to 1 by Pe times d square T by dx square plus d square T by dy square.
Even though this prefecture, may be small because the peclet number is large, if the
derivatives is large, the product of these two could still be comparable to the convection
term, the product of these two could still be comparable to the convection term.
Note, that I when did the scaling here, when I did the scaling here, I implicitly assume
that y star is equal to y divided by L. That means, the line scale for variation of temperature
in the y direction is equal to L. So, when I did the scaling in this manner, I implicitly
assume that L is the line scale of which the temperature varies in the y direction. On
that basis, I did my scaling and then, I found out that convection was large compared to
diffusion. And therefore, I try to neglect the diffusion term. However, if the length
scale for diffusion is smaller than L, the length scale for diffusion is smaller than
L then the gradient will be larger, because the derivative goes as temperature difference
divided by the line scale of variation. So, the length scale is smaller. The gradient
could be much larger and if the length scale is sufficiently small, I could still get a
balance between convection and diffusion, even though the peclet number is based on
large. I could have a smaller length scale of which their diffusion such that the gradient
sufficiently large, such that the balance between convection and diffusion.
That is what is actually happening in high peclet number flows. Ready near the surface
diffusion has to exist. Because, that is the only way the flow is to get heated. But however,
if I go line scale comparable to L from the surface, my balance between convection and
diffusion tells me convection has to be dominant. If I go a distance L away from the surface,
diffusion could still be important within much smaller length scale, because convection
is so fast, the heat can penetrate very far into the fluid. But, diffusion was still be
important over a much smaller length scale. And that length scale is determinant in such
a way or has a sufficient value that, over a distance comparable to that, there is balance
between convection and diffusion. So, how I am going to solve this problem? I said physically,
the reason that we are not able to get balance is because, we have assumed that the length
scale for variation in the y direction is capital L. If the length scale is variation
in the y direction smaller than capital L, there might be balance between convection
and diffusion. What should that length scale be?
So, what I will do now is to just write down, scaling x star is equal to x by L. This is
usual, this is the length scale for variation of temperature, along the x direction, because
the heated section has length capital L. However, in the y direction I will postulate smaller,
in the length scale. A smaller length scale, smaller. And then go back and scale my equations
once again. So, my original equation was gamma dot y partial T by partial x is equal to D
into partial square T by partial x square plus partial square T by partial y square. There should be thermal
diffusibility alpha. And with this scaling as well as T star is equal to T minus T naught
by T1 minus T naught. This becomes gamma dot L y star by capital L partial T by partial
x is equal to alpha into 1 over L square. So, this now is my convection diffusion equation
with y scaled by, as unknown scale length smaller. Note that, within this equation in
the diffusion terms, this one contains prefactor, 1 over small l square. The first terms contains
the prefactor 1 over capital L square. It postulated that small l, was small compared
to capital L. That means, that this diffusion term in the y direction is large, compared
to the diffusion term x direction. Because, the diffusion term in the y direction goes
has small l square, the diffusion term in the x direction goes as over capital L square.
That means, that this second coefficient term, in the y direction, the cross stain direction
is large compared to the steam vise direction. So, I divide throughout by alpha by L the
whole square, if I divide throughout whole alpha by L whole square, rather in the limit
of high peclet number I should divide by this term, divide throughout.
The equation becomes y star dT by dx is equal to alpha L by l gamma dot into 1 over l partial
square T by partial y square plus 1 over L square. Alternatively, I could write this
as alpha L by l cube gamma dot into. So, this then is my convection diffusion equation.
Note, that this is small number, this is a small number, because I postulated that small
l compared to capital L. Therefore, I can neglect stream vise diffusion comparison to
cross strain diffusion. If I do that, what I get is that y star partial T by partial
x is equal to alpha L by l cube gamma dot partial square T by partial y square. Now,
this is the convection diffusion equation. And if convection and diffusion have to be
comparable in the high peclet number limit, it means that this term here, has to be of
order 1 in the high peclet number limit. If convection and diffusion are to be comparable
in the limit of high peclet number, I require that this term has to be order 1. That means,
that l cube gamma dot by alpha L is 1, which means that l by L the whole cube is equal
to alpha by gamma dot L square. Now, alpha by gamma dot by L square was the
original peclet number that I had inverse. Because, if we recall, we defined peclet number
has equal to gamma dot L square by alpha for this problem. So, therefore, I require that
l by L is equal to Pe power minus one-third. So, what this is telling me is that, in the
limit has the peclet number becomes large, there is a length scale, length; very near
the surface small l over which diffusion is comparable to convection and that value of
small l decreases Pe power minus one third in the limit has peclet number becomes large.
I said that this term has to be just order 1. But, however, I can set it equal to 1,
without loss of geniality, because it is just a length scale, which I obtaining by scaling.
I can set equal to any constant value, it will change the equations in terms of the
length scale L, what I get a final solution for the physical problem, there will be no
dependence on this constant. So, without loss of gentility, I said just
the constant is equal to1. This is what we called boundary layer thickness, the thickness
of the region over which there is balance between convection and diffusion. Even, when
the peclet number is large for this reason that I said even, when the peclet number is
large, even though it appears the convection is dominant, is large compared to diffusion,
there is still going to be the region very close to a surface where diffusion is present.
And it is comparable to convection. That is because the transport from a surface cannot
takes place due to convection, because there is no velocity normal to the surface. Diffusion
on the other hand is an isotopic process. It is due to the fluctuating velocity of these
molecules which have equal magnitudes in all directions. And therefore, they transport
mass, momentum, energy equally in all directions. Therefore, very close to the surfaces even
though convection is not present, there is still diffusion, and that diffusion is what
causes transfer from the surfaces itself. And the variation of concentration very close
to the surface has a length scale L, which is determined from the requirement, that convection
and diffusion have to be comparable very close to the surfaces. This requirement itself,
what is the value of the length scale. In the limit as Pe goes to infinity, the peclet
number becomes large, this length scale goes as Pe power minus one-third, in the limit
has peclet number becomes large. And as we saw just now in this equation, stream vise
diffusion is always small compared to the cross stream diffusion, very close to the
surface. Because, the length scale for cross strain diffusion is the small l, which goes
Pe power minus one-third. The length scale for steam vise diffusion is the macroscopic
scale capital L itself.
Therefore, for this particular problem the convection diffusion equation reduces to gamma
dot y dT by dx is equal to alpha d square T by dy square. For this particular problem,
this is the equation. We can neglect the steam vise diffusion in comparison to the cross
vise diffusion. Now, we obtained based upon scaling, this as the simplified equation.
How do we solve this? This is still a partial differential equation contains variations
in both x and y. The solution procedure for this simplified equations is, we can formulate a solution
procedure based upon are physical understanding of this problem.
So, I have the section and then I have a heated section here. So, this is y and this is x.
And this heated section has some length L. The flow is incident on this heated section.
So, T is equal to T naught, or T star is equal to 0. T star is equal to 0 and T star is equal
to 1. And we are considering the limit, where convection is dominant comparison to diffusion.
And on that basis, that there will be some boundary layer length here, length of small
l, which goes as L times Pe power minus one-third, over which there is balance between convection
and diffusion. However, if you consider for example, any
particular location, if you consider any particular location x, we consider any particular location
x, I have diffusion coming out of the surface, diffusion of heat is coming out of the surface,
and there is convection. Now, convection is taking the energy downstream. convection is
taking the energy downstream. We are considering, in this case convection is large compared
to diffusion based upon the length L. So, convection is taking energy downstream, the
diffusion from the surface. That means, at a given position x the temperature should
not depend upon the total length L because, the total length L is downstream of the position
x. The temperature at given location x cannot depend upon the total length L downstream
of that location. It should depend only upon the length of the heated section up to that
particular location x. It should only depend upon the length of the heated section up to
that particular location x, that means, that at the only length scale which the temperature
at given location x should depend is on the distance x from the start of the heated section
itself not the total length of the heated section, because the total the remainder of
these heated section downstream of these position and flow is carrying the heat downstream.
So, that can be no diffusion in the stream vise direction which basically, enables the
temperature at this location x to feel the downstream locations.
So, if x is the only length scale in the problem and the boundary layer thickness l at a given
location x divided by x has to be equal to the peclet number based upon x to the minus
one-third. Which is equal to alpha by gamma dot x square power one-third. So, the only
length scale in the problem is x. So, I will just substituted x for L in this particular
case assumed that the boundary layer thickness L is the function of x. This will enable us to get a similarity solution
because of the length scale, the y direction is L, then I can define a scaled similarity
variable has y by L of x. And L of x is equal to alpha x by gamma dot power one-third, if
I can define this as y by alpha x by gamma dot power one-third.
This was obtained just basically from the consideration that at a given location x the
temperature should depend only upon the distance from the start of the heating section, not
the total length of the plate itself. So, I cannot scale my x, my y coordinate the boundary
layer thickness by capital L. Because, at a given location x, it should not depend upon
the total length L. It should depend only upon distance x from the upstream section.
So, my boundary layer thickness L of x has to be alpha x by gamma dot power one-third.
If this is the boundary layer thickness, then I just divide y by that scaled variable because
the characteristic length scale in the y direction is this small l. So, I can define eta is equal
to y by alpha x by gamma dot power one-third. This is the similarity variable, which reduce
the problem from two dimensions into one dimension. If you recall the impulsively impulsively
started plate, there we had obtained the similarity solution based upon dimensional analysis the
fact that, there is no length scale or time scale in the problem. Therefore, I can get
only one similarity variable, here we are doing based upon physical insect.
So, now, I have to substitute in this convection diffusion equation. I have the convection
diffusion equation gamma dot y partial T by partial x is equal to alpha partial square
T by partial y square. So, in this equation I have to substitute y and x in terms of eta.
So, I will get
partial T by partial y is equal to partial T by partial eta times partial eta by partial
y is equal to 1 over alpha x by gamma dot power one-third partial T by partial eta d
square T by d y square is in similar manner 1 over alpha x by gamma dot power two-thirds.
And dT by dx is equal to dT by d eta times d eta by dx is equal to y by power one-third,
x power minus one-third. The derivatives this will give you minus 1 by 3 times x power minus
4 by 3. This finally, has to be substituted into this differential equation. And then
we have to get solutions for this differential equations. If what we done here is correct,
then we substitute this into that differential equation, I will end up with this equation
in terms of eta alone. It should not depend separate x and y. The final equation should
depend upon eta alone.
So, what we get substitute this? Get gamma dot y into minus y by 3 x alpha x by gamma
dot power one-third dT by d eta is equal to alpha into divided by alpha x gamma dot power
two-thirds d square T by d eta square. So, I can simplify this a little bit by using
the simplification y equal to eta into alpha x by gamma dot power one-third. So, I will
get gamma dot eta square into alpha x by gamma dot power one-third by 3 x into alpha x by
gamma dot power one-third dT by d eta is equal to alpha by power two-thirds d square T by
d eta square. And you can easily verify that all of these terms will cancel out.
Finally, to give you minus eta square dT by d eta is equal to d square T by d eta square.
So, this then is the final solution for the equation in terms of eta. As we expected the
scaled the y coordinate by that length L, which is based upon the x, x distance. The
distance from the upstream edge, then you will finally end up the solution in which,
it does not individually depend upon x, y, gamma dot and alpha. It depends only upon
the similarity variable eta. So, this gives us the solution the temperature field in terms
of the similarity variable alone and does not upon separately on x and alpha, y and
x, but only on the similarity variable alone. So, now we have equations in terms of the
similarity variable eta. We have to reformulate the boundary condition as well in terms of
similarity variable eta. The original boundary conditions, at y is equal to 0, T star is
equal to 1 T star is equal to 1, at y is equal to 0 implies that eta is equal to 0, because
eta is equal y by alpha x by gamma dot by alpha power one-third. As y goes to infinity
T star is equal to 0, y going to infinity implies that, this implies that eta is going
to infinity. And at x is equal to 0 for y greater than 0, T star is equal to 0. And
you can see from this, that x is equal to 0, basically means that eta goes to infinity.
x is equal to 0 basically means that eta goes to infinity. So, at x is equal to 0. So, you
can both of these boundary conditions one boundary condition in the y coordinate as
y goes to infinity and the other, x coordinate at x is equal to 0 reduced the same boundary
condition, expressed in terms of eta. These two reduces exactly the same, when expressed
in terms of eta. That is expected. Originally, I have the first order differential
equation in x and second order differential equation in y, shown in on red, right on top
there. The first order in x and second order in y, the two conditions in y and x. When
I reduce it transform, I got only one second order differential equation in eta x L there
is only two boundary conditions. That means, that one of the boundary conditions y as well
as the initial condition in x has to reduce to same when expressed in terms of eta. So,
I can solve this equation minus eta square partial T by partial eta is equal to partial
square T by partial eta square, integrated once to get partial T by partial eta is equal
to e power minus eta cube by 3. I integrated once to get partial T by partial eta is equal
to C. I am sorry, some constant C1 minus eta cube by 3.
And I can integrate this as second time, to get T star is equal to integral from 0 to
some value of eta. eta prime exponent of minus eta prime cube by 3 plus some other constant
C2. So, this is the final solution. This integral cannot be evaluated exactly because of the,
because, we have definite integral here. Because, we cannot integrate minus e power cube exactly.
And then we apply the boundary conditions. T star is equal to 0 as eta goes to infinity
and T star is equal to 1 at eta is equal to 0, in order to determine, specify constants
C1 and C2. So, the constants turnout to be this final solution, after imposing the boundary
conditions turns out to be eta prime power eta prime minus power 2 by 3 divided by integral
0 to infinity. You can easily verify that this solution is 1 at eta is equal to 0 and
it is 0 as eta goes to infinity. This is the final solution for the temperature field.
In terms of the variable eta where eta is equal to y by alpha x by gamma dot power one-third.
There is the final expression in terms of this eta.
How can we use this to calculate the heat flux? The total amount of heat coming out
of the heated surface? Locally the heat flux q y is equal to minus k dT by d y. I have
to express this in terms of eta, if you recall dT by d y is equal to 1 by alpha x gamma dot
to power one-third times dT by eta. The heat flux coming out of the surface y is equal
to 0. So, this will be equal to minus k by alpha x by gamma dot power one-third dT by
d eta at eta is equal to 0. And dT by d eta at eta is equal to 0 can be quite easily calculated
from this solution here.
And d T by d eta is just equal to minus, so this q y will be equal to
minus k by alpha x by gamma dot power one-third into minus 1 by integral 0 to infinity d eta
prime, e power minus eta prime cube by 3 into T1 minus T naught. Because T is equal to T1
minus T naught, T minus T naught minus T 1 minus T naught times T star. This is the final
expression, we get for the heat flux. So, this is equal to k into T 1 minus T naught
by alpha x by gamma dot power one-third times. This function can be integrated exactly gamma
1 by 3 by 3 power two-thirds. That is the final solution heated flux from the surfaces.
The average heat coming out of the surface, the average heat coming out from the surface
is the average over an entire length 0 to L dx times q y at this location x. And this
can be evaluated analytically, in order to get the average heat that is coming out from
the surfaces. From this, we will now go calculate the nusselt number coordination for the heat
transfer from the surface in the limit of high peclet number. When we did the initial
lectures, I showed you that the nusselt number, in the case of convicted dominant flows at
high peclet number goes Pe power plus one-third. You have done here an analytical solution
for the flow past a flat surface peclet numbers. And we got a solution in terms of similarity
variable. I will now use this, how we get the high nusselt number coordination. So,
I will first complete the solution of flow past a flat surface peclet numbers and then
got a solution in terms of similarity variable. I will now use this to show you, how we get
a high peclet number coordination, complete the solution of flow past the flat plate in
the next lecture. And then, I will discuss the flow past the spherical particle and evaluate
the nusselt number coordination. So, slowly we are recovering all original results that
we had as a empirical correlation is the same thing deriving here, by exact calculation.
Using certain approximations the low nusselt number limit, I am sorry the peclet number
limit and the high peclet number limit. We will continue this in next lecture. Will see
you then.