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Welcome to the sixth lecture in this lecture series on fundamentals of transport processes.
As promised, this lecture is going to be about diffusion.
Fundamentals of transport processes - two mechanisms - convection and diffusion. I had
earlier explained to you, why diffusion is so important.
Let us just go over that once again. In the problem of the reactor, solid catalyzed reactor,
the transport of mass of the reactants to the surface and the transport of products
from the surface, the transport of heat and without that transport, the reaction is not
going to take place. If you look closely at the surface itself,
the fluid flow does not penetrate the surface because the surface is an impenetrable surface.
Therefore, the fluid flow has to be tangential to the surface; a fluid flow tangential to
the surface is not going to transport mass or heat to the surface. It is going to transport
only along the surface. For the mass to go to the surface, you need have transport that
is perpendicular to the surface; that transport can take place only due to diffusion.
So, ultimately, there has to be diffusive transport for the material to be transported
to the surface. In the case of the heat exchanger, there is convection of hot fluid into the
heat exchanger. So, there is heat coming in with that hot fluid by convection, but that
heat has to go across the surface to the coolant. Across the surface, there is no fluid flowing.
Fluid flow is only along the surface. Therefore, transport across the surface ultimately has
to take place due to diffusion.
Similarly, in the problem of spray drier that I had talked about, the mass that is transported
from the catalyst surface to the outside or the heat that is transported into the droplet,
both of these had to be transported due to diffusion because there is no fluid penetrating
the surface. So, there is no fluid velocity perpendicular to the surface. Transport perpendicular
to the surface has to happen due to diffusion. Same is the case of momentum transfer. There
is no fluid going in or going out. So, it carries no momentum with it. The only force
that is exerted is due to the drag force on the surface, the shear stress that is exerted
on the surface as well as the pressure forces. This transport of momentum by shear stresses
has to happen due to diffusion. So, ultimately, regardless of whether convection is dominant
or diffusion is dominant, if you go very close to the surface, the transport ultimately has
to take place due to diffusion.
So, what is diffusion? Diffusion is the transport of material due to the fluctuating motion
of the molecules without any mean velocity. Convection takes place when there is a mean
motion of the fluid; diffusion takes place due to the fluctuating motion without any
net mean motion in the fluid. Let us take a simple example of mass diffusion.
Let us assume that we had 2 cylinders with a stop cork in between it. So, this one was filled with molecules of gas; this one
was filled with molecules of the same gas, but with a little bit of aromatic molecules
added in here and there. Initially, the 2 compartments are separated,
but at some time we remove the separation between the 2 compartments. So, if I remove
the separation between the 2 compartments, these molecules are going to be in fluctuating
motion. So, if I look at some local location over here and I expand it out.
I have these black molecules here and I have black molecules below as well with a few red
molecules thrown in and all of these molecules have fluctuating velocities.
As time progresses, these red molecules are going to diffuse around and inevitably some
red molecules are going to cross the surface and go from above to below. There are no red
molecules above; so, there is no equal transfer down and because of that, there is going to
be net transfer of molecules from below to above
So, if I plotted the concentration field as the function of distance, for the red molecules,
initially there are no red molecules above, so, I will get concentration that looks something
like this. There is concentration of red molecules below, but there is no concentration above.
As time progresses, these molecules will slowly diffuse across. I will get a field that looks
something like this and if I wait for a very long time, the system gets completely mixed
up and the concentration is the same everywhere. As long as there is a concentration difference
between below and above, there is going to be a flux of molecules. Note there is no mean
motion of the molecules themselves; molecules themselves are only in a state of random motion.
So, there is no change in the center of mass. However, there is a flux of molecules because
of the fluctuating velocity of the molecules coupled with the difference in concentration.
When you reach the final state, where the concentration is the same throughout, there
is once again no flux of molecules. So, flux of molecules requires a change in concentration
across the length. At a molecular level, why does diffusion take
place? The simplest example to consider is that for a gas. So, in this particular case,
let us look at the molecular transport. All of the molecules are having some random fluctuating
velocity and because of that there is a constant motion of the molecules across the surface.
Now, in a gas, the molecules move a distance comparable to the mean free path between successive
collections. So, molecules that are going above from below are going from a distance
comparable to the mean free path below the surface whereas, the molecules that are coming
from above downwards are going some distance comparable to the mean free path from above
the surface.
If I take this as the location where y is equal to 0, the molecules coming from below
are coming from a distance approximately y minus some lamda times a constant, a distance
comparable to the mean free path below the surface. Molecules going from above to below
are coming from a distance comparable to the mean free path above the surface.
There is a difference in the concentration between these two. The concentration of red
molecules is higher below and lower above. Therefore, you have more red molecules going
above and less red molecules coming below and that ultimately is what is causing diffusion.
So, there are two things: one is fluctuating velocity of molecules and the other is the
concentration of the molecules at the point from which they come, the point at which they
originate in the case of gases.
So, let us look at it more closely. This is my y coordinate. I have given a location and
the concentration has some variation about this location. I should draw in the other
way. concentration has some variation about that location.
Now, the molecules that are going upward are coming from some distance of the order of
mean free path below the surface. We can do more detailed calculations and find out exactly
how much they are coming? It turns out that they come from a distance minus 2 by 3 times
the mean free path lamda. So, the molecules from here are the ones that
ultimately cross the surface. What is the flux of molecules going upwards? The flux
of molecules is going to be equal to the concentration of molecules at y minus 2 by 3 lamda. This
factor 2 by 3 comes from more detailed calculation which we shall not concern ourselves over
here. We will just assume that it is some constant factor; it is comparable to the mean
free path of the same order of the magnitude as the mean free path. So, that gives me concentration
times the root mean square velocity of the molecules.
Molecules are coming from a distance comparable to the mean free path below. So, the flux
will be this concentration times v rms, not exactly v rms because there will be some multiplication
factor. So, it will have some factor a in front of it times C times at y minus 2 by
3 lamda times v rms. Once again, more detailed calculations show that this factor is actually
one-fourth. I will not able to do the calculation to show that right now, but for the present we will
just assume that they are coming the flux is proportional to the concentration times
the root mean square velocity times the constant. That concentration is not at the location
y, but at a location below the surface. The distance between the location and the surface
is comparable to the mean free path and more detailed calculations show that y is equal
to 2 by 3 lambda. So, that is the flux going from below to above.
What about the flux going from above to below? The flux of molecules going from above to
below come from a distance of the order of mean free path above the surface.
Therefore, the flux going downwards will be equal to one-fourth C into y plus 2 by 3 lamda
times the root mean square velocity. Total flux - what goes above minus what goes below,
if the y direction is directed upwards. If you assume that this is positive y upwards,
y is upwards, positive in the upward direction, then the total flux j in the y direction is
equal to j plus minus j minus - what goes up minus what comes down.
So, this will be equal to one by fourth v rms into C at y minus 2 by 3 lamda minus C
at y plus 2 by 3 lamda. Now, continuum description as I discussed in the last lecture is valid
only when the macroscopic scale is large compared to the microscopic scale, the mean free path
in this case. So, when the macroscopic scale is large compared
to the microscopic scale, then this distance which we are interested in, the variation
over distances, the distance range over which we want to find out variations is large compared
to the mean free path. In that case, I can do a Taylor series expansion of the concentration
about the location y. It is C at y minus 2 by 3 lambda is equal to C at y minus 2 by
3 lambda into dC by dy y plus higher order terms.
C at y plus 2 by 3 lamda is equal to C at y plus 2 by 3 lambda dc by dy y plus other
terms. So, locally over distances comparable to the mean free path, if you assume the concentrated
field is approximately linear, the variations and concentration takes place over much larger
distances. So, locally the concentration field looks linear over distances comparable to
the mean free path. Then we can use this Taylor series expansion and with this, the total
flux j y take this and dark j y will be equal to one by fourth v rms into C at y minus 2
by 3 lambda dC by dy y minus C at y minus 2 by 3 lamda times dC by dy.
Finally, this will give me minus 1 by 3 v rms lamda times dc by dy. So, this is the
flux equation. Compare with the flux equation that we have got in the previous lecture.
In the previous lecture, we got j y is equal to minus D times dC by dy, D is the coefficient
in front of the gradient, in front of the concentration gradient; concentration gradient
here, flux here, the co efficient in between this the diffusion coefficient.
Therefore, the diffusion coefficient in this case, will be one-third v rms times lamda.
Two things: one the flux depends upon the variation of the concentration with respect
to position. Therefore, there is a diffusive flux only when there is a variation of concentration
with respect position; this is called the gradient of the concentration field.
I told you that the Fick’s law relates the mass density to the rate of change with position
of the mass density - the concentration. So, the proportionality constant in that case
has to have dimension of length square per unit time. You can easily see that v rms times
lamda has dimensions of length square per unit time.
Velocity has the dimensions in length per unit time; lamda has dimensions of length.
So, this is the diffusion coefficient calculated from microscopic prospective with an approximation.
We had made 2 approximations here. One was that flux was equal to this factor of one-fourth
times v rms and the second approximation related to the distance from where the molecules came.
A more exact calculation can be done and the only thing that changes is this coefficient
in front here. We will discuss this coefficient little later,
but the dependence on the root mean square velocity and the mean free path in a gas remains
exactly the same. So, in all cases, for all gases, the diffusion coefficient has to be
proportional to v rms times lambda because diffusion occurs due to the fluctuating motion
of the molecules. That fluctuating motion depends upon the root mean square velocity
and the diffusion takes place over a distance comparable to the mean free path because that
is the microscopic scale. So, it is going to be sensitive to the variations
in concentration over a distance comparable to the mean free path. The change in concentration
over a distance comparable to the mean free path is going to be equal lamda times dC by
dy and that is why the diffusion coefficient has this form.
Next, so, how do we estimate this diffusion coefficient? We have to know what is v rms
and we have to know what is lamda. We will go through that for a gas in order to justify
why the diffusion coefficient that are actually measured in experiments have the values that
they have.
So, in kinetic theory of gases, you can define mean square velocity in different ways. v
rms is defined as square root of 3 k T by m; 3 k T - this comes from the equipartition
of the energy relation, half m v square is equal to 3 by 2 k T.
So, this is approximately the root mean square velocity of the molecules in a gas. I can
define a different velocity, which is the mean. This comes out to be is equal to square
root of 8 k T by pi m. This averaging is done in a spherical coordination system in the
velocity space that need not concern us too much over here. The constant in both cases
do not turn out to be very different. So, what is the root mean square velocity
for normal gases? Let us try to estimate that. k is approximately 1.38 times 10 power mines
23 joules per Kelvin. If we take T as approximately 300 kelvin at room temperature, then k T will
be approximately 4 into 10 power minus 21 joules; that is an estimate of k T, the energy
scale in the system at room temperature approximately. What about the molecular mass? The molecular
mass of course, varies. This is the mass of a molecule. For example, if you take the example
of hydrogen, the mass of 1 mole is 2 grams; 1 gram mole is 2 grams. Therefore, the mass
is equal to 2 into 10 power minus 3 kilogram for 6.023 into 10 power 23 molecules.
From that, you can get the molecular mass is equal to 2 into 10 power minus 3 by 6.023
into 10 power minus 23 kilograms, which according to my calculation turns out to be about 3.32
into 10 power minus 27 kilograms. So, this is the mass of 1 molecule. On that basis,
I can find out what is v rms; it is equal to square root of 3 k T by m. With these figures,
my calculation shows me that this is about 1.29 into 10 power 3 meters per second.
So, this is the fluctuating velocity of a hydrogen molecule in a hydrogen gas. If you
take for oxygen molecule, for example, for oxygen, mass is equal to 32 into 10 power
minus 3 kilograms. So, the mass is 16 times that for hydrogen. Therefore, the rms velocity
will be about 1 by 4 times that of hydrogen and my calculation shows me about 321 meters
per second. So, this is the magnitude of the fluctuating
velocity of the gas molecules. We can see that they are quit large. A thousand meters
per second for hydrogen, more than a kilometer a second for oxygen, it is about 321 for nitrogen
and air, it will be roughly about the same the numbers, will be approximately the same
because the molecular mass of nitrogen is not very different from oxygen.
Your will also notice that these velocities are also approximately equal to the speed
of the sound. In air, we know that the speed of sound is 330 meters per second and the
rms velocity turns out about 321; in hydrogen, the speed of sound is about 1290 meters per
second and that is what we have got for v rms. So, the molecules are fluctuating with
velocities that are comparable with the speed of sound. So, the rms velocity of molecules
in a gas is comparable to the speed of sound and the estimates are between 100 to a 1000
meters per second approximately.
So, the next thing that we need is the mean free path. Now, this is a little more complicated
and I will try to go slowly through it, but it is a useful calculation because it gives
us some physical insight into why the numbers turn out to be the way they are.
So, a typical molecule will undergo collision with other molecules and will go along a straight
line path for some time and it will undergo another collision with some other molecule.
Now, this molecule will travel in straight line between successive collisions. So, as
it is traveling in the straight line, it sweeps out a cylindrical volume.
So, as these molecules travel in a straight line, it sweeps out a cylindrical volume.
This molecular diameter is d; it sweeps out a cylindrical volume of radius equal to d,
such that if the center of second molecule is lying within this cylinder, there will
be a collision between 2 molecules. If the center of the second molecule is lying within
the cylinder, there will be a collision between the 2 molecules.
So, as it is traveling with its root mean square velocity, it is sweeping out this distance.
If it travels some length L, the volume of cylinder swept out is going to be equal to pi d square
times L and of course, as it sweeps out a longer and longer cylinder, the probability
that it will collide is going to increase and why is that going to increase? It is because
the probability of finding the second molecule within the cylinder increases. Therefore,
let us estimate what is the probability of finding the second molecule within a cylinder
of the length L. The probability of finding a second molecule
is going to be equal to the number density of molecules times pi d square L and n is
number density of molecules - number per unit volume - pi d square L is a volume. So, the
products of these two becomes just a number So, as the number of second molecules is increases,
there is a greater and greater chance of collision. The probability that the molecule will all
most certainly collide happens to be 1, when L is equal to the mean free path because it
cannot go much beyond the mean free path. It is very rare for a molecule to go very
much larger than the mean free path before it collides.
Therefore, it is going to reach, it is going to undergo a collision and this probability
becomes approximately 1, when this distance L is equal to the mean free path lamda. Therefore,
I have n pi d square times lamda is approximately 1. So, it is going to collide it is going
to cover its mean free path when this is approximately 1. Therefore, lamda has to be approximately
one by pi nd square. So, this is a simple estimate that was based
upon single molecule traveling in a straight line between successive collisions. You can
do a more detailed calculation and you will get the mean free path is 1 over root 2 pi
nd square.
So, that is the mean free path. Let us try to estimate what is mean free path? The number
of molecules per unit volume is equal to p by k T, number of molecules per unit volume
from p v is equal to n k T. So, if you take atmospheric pressure, the pressure is approximately
1 into 10 power five Newton’s per meter square divided by k T which is 4 into 10 power
minus 21 joules. You can easily see that these has dimensions
of one over volume and if I carry out this calculation, this number density becomes approximately
2.5 times 10 power 25 molecules per meter cube at standard temperature and pressure.
So, this is the number density of molecules. The mean free path is going to be equal to
1 by root 2 pi nd square. For hydrogen, d is 1.38 angstroms which is approximately 1.38
times 10 power minus 10 meters and using this in this formula, I get the mean free path
is equal to 0.5 into 10 power minus 6 meters or approximately 0.5 microns. One micron is
1000 of the centimeter. So, this is approximately 1 micron is about 1000 centimeter.
For oxygen and nitrogen, I think diameter is approximately 3.7 to 3.8 angstroms. If
you put this into this formula, I will get lamda is approximately equal to 6 into 10
power minus 8 meters. So, this is approximately 16 nanometers and you can easily see that
the diameter is about three angstroms and the mean free path is about 6 nanometer and
that the mean free path is approximately 100 times the molecular diameters, in the case
of gases. So, this gives a range to the mean free path
is between about 0.5 microns to 10 power minus 7 to 10 power minus 8. So, diffusion coefficient
is equal to some constant; in this case, we got one-third v rms times lamda. For hydrogen,
v rms was about 12100 meters per second and I am getting about 0.5 micron per lambda and
on this basis, I can estimate the diffusion coefficient as approximately 6 times 10 power
minus 4 meter square per second for hydrogen. This is approximately 2 times 10 power minus
5 meter square per second for oxygen and nitrogen. So, this is the range of diffusion coefficient for most common gases. It is approximately
Diffusion coefficient in all cases is approximately 10 power minus 5 meter square per second.
If you look for example, look for values in literature, Kastler reports that H 2 and He
diffusion coefficient is about 1.132 into 10 power minus 4 meter square per second,
while for oxygen, it is about 1.8 times 10 power minus 5 meter square per second.
So, this is the range of diffusion coefficient for most common gases and the reason that
the diffusion coefficient has this range is because the mean free path goes from about
10 power minus 7 to 10 power minus 8 meters and the velocity varies between 100 and 1000
meter square per second. In fact, you can do more detailed calculations to get exact
results for diffusion coefficient for let us say spherical rigid molecules.
This turn out to be of the form D is equal to 3 by 8 nd square into k T by pi m power half. This is for a
single component, where the number density is where the masses of the 2 molecules are
not very different and if you have multi component diffusion - 2 species, this will be 3 by 8
n 1 2 d 1 2 square into k T m 1 plus m 2 by pi m 1 m 2 power half, where d 1 2 is equal
to d 1 plus d 2 by 2 and n 1 2 is equal to square root of n 1 n 2. So, this is for multi
component diffusion. This explains why the diffusion coefficient
have the values they do for gases. How about for liquids? So, for gases, I said the diffusion
coefficient is of the order of 10 power minus 5 meter square per second. For liquids, the
thermal velocity is the same. The v rms, liquid’s v rms is still equal
to 3 k T by m power half. The thermal velocity is the same whether it is gases or liquids.
Only thing is in liquids, the distance between the molecules is small. Therefore, the molecules
cannot move very far between successive interaction; that is the only difference between liquids
and gases. On this basis, the thermal velocities are
the same, the mean free path in the liquid - distance between molecules, is about 100
times smaller than the mean free path in a gas as I just told you. Therefore, you would
expect the diffusion coefficient also to be 100 times smaller. So, you would expect diffusion
in liquids to be approximately 1 by 100 the diffusion coefficient for gases.
Turns out that this is not true; this equation it turns out is not true and the reason is
the high density of molecules in a liquid. So, if I had molecules in a liquid, the motion
of 1 molecule in any particular direction requires that all other molecules move out
of the way. So, diffusion in liquids is basically what
is called cooperative processes. The motion of 1 molecule requires other molecules to
move out. For most of the times, the molecules in a liquid are rattling within the cage that
is made of the neighbouring molecules and occasionally there is a rearrangement, where
one molecule moves out because the neighbouring molecules are moved in such a way as to permit
the passage of the molecules. Transport of mass requires the motion of molecules,
the physical motion of molecules and for this reason diffusion in liquid is much slower
process. In fact, if you calculate the diffusion coefficient in liquids, for small molecules,
it is approximately 10 power minus 9 meters square per second which four orders of magnitude
smaller than that in a gas. For example, for hydrogen and water, the diffusion
coefficient is 4 into 10 power minus 9. For larger molecules, if you had larger polymer
molecules, protein and so on, this could be as low as 10 power minus 11 to 10 power minus
13 meters square per second, for large. So, the diffusion coefficient in liquids is
an exceedingly slow process because it involves the cooperative motion of many molecules and
the diffusion coefficient in liquids is difficult to estimate just based on the mean free path
consideration alone, but there is alternate ways to estimate that. One equation is what
is called Stokes Einstein equation - k T by 3 pi mu times the molecules diameter. k T
is the same k T, the thermal energy that we had before, mu is the viscosity of the liquid
and D is the diameter of the diffusing particle; this called Strokes Einstein relation.
This relation is strictly speaking valid only for large collidal particles in a liquid,
There is a requirement in deriving this relation that this diameter D has to be large compared
to the molecular diameter so that the relaxation time of the large particle is much larger
than the time scale on which these small molecules impact on the particle, but this can still
be used to give an estimate of what diffusion coefficient of liquid should be
Finally, there is one other point to be noted here. In gases, D is equal to some constant,
let us call that as a, times lamda times v rms, where lambda is equal to 1 by root 2
pi nd square and v rms is equal to square root of k T by m. That means, in gases, the
diffusion coefficient increases with temperature. The reason is because as you increase the
temperature, the fluctuating velocity of the molecules increases and because of that the
diffusion coefficient increases. The diffusion coefficient also increases as the number density
is decreased; as n decreases, the diffusion coefficient increases.
That is because molecules travel longer distances between successive interactions. So, at a
given surface, as you decrease the number density, the molecules of that point coming
from much lower, where the concentration difference is much larger. That is the reason the diffusion
coefficient increases as you decrease the number density and it increases proportional
to T power half as the temperature is increased.
In liquids, on the other hand, you can see from the Stokes Einstein relation, the diffusion
coefficient has this form. So, it depends both upon the temperature as well as the viscosity.
Typically, the diffusion coefficient will decrease as the temperature is increased.
Even though the temperature increases, the viscosity decreases. The diffusion coefficient
increases as the temperature increases because the viscosity is decreasing fast and the temperature
is increasing. When the viscosity decreases with increase in temperature, the diffusion
coefficient will increase with the increase in temperature.
In this case, the power is actually larger than one because the temperature is increasing
and the viscosity is decreasing. So, in both cases the diffusion coefficient does increase
with temperature, but for different reasons in both cases.
So, that gives us some physical insight into the diffusion of mass. Now, let us look at
the diffusion of momentum. That is I am trying to get a relationship between the flux of
momentum, which is the shear stress and the change in velocity across the surface. So,
let us consider shear flow in which the velocity, this is the y direction, the velocity u x
is varying along with position y. So, this is the location y is equal to 0 and I have
molecules here and all of this have fluctuating velocities. However, there is also a mean
velocity. So, the molecules on top are moving with a
small mean velocity towards the right; these molecules below are moving with a small mean
velocity towards the left. The implicit assumption in all of these cases is that, the mean velocity
is small compared to the fluctuating velocity of the molecules. As I told you earlier, the
fluctuating velocity of the molecules is comparable to the speed of sound in the medium.
So, the assumption in all of these cases is that the mean velocity is small compared to
the speed of sound. The ratio of mean velocity in speed of sound is called Mach number and
we deal exclusively with low Mach number flows, where the mean velocity is small compared
to the speed of sound. So, if you take the volume below the surface
and try to write an equation for the rate of change of momentum. Note that momentum
has 3 components, but in this particular case you are assuming that the mean velocity is
non-zero only in the x direction. So, momentum only has an x component; there is no y, z
component. However there is a fluctuating velocity of the molecules in all directions.
In particular, there is a fluctuating velocity of molecules in both x and y directions and
that will prove to be important when we discuss diffusion.
So, now, what is the rate of change of momentum within this volume? When a molecule goes from
below to above, the momentum of this volume decreases. When a molecule comes from above
to below, the momentum of this volume increases. Now, we have to find out the flux of momentum
from below to above and from above to below. Flux of momentum from below to above, I will
call this as j y momentum flux and this momentum flux is going in the positive direction. This
momentum flux is going to be equal to the average momentum of the molecules below times
the number of molecules traveling per unit time per unit area.
So, the average momentum of the molecules below this surface is going to be equal to
number of molecules times mass of a molecule times v x at y minus 2 by 3 lamda is the value.
So, this is the mean momentum. So I have to be careful here. I should use u x n into u x y minus lamda times the flux
of molecules. So, this is the momentum of molecules below
times the flux of molecules, which is approximately v rms times some undetermined constant a.
In the previous case, it was 1 by 4; you do not worry too much about what the exact numerical
value of the constant is. So, this is the momentum that is going from
below to above. The momentum coming from above to below is going to be similar except that
this is equal to nm times u x at y plus 2 by 3 lamda into v rms times proportionality
constant.
So, now, what is the rate of change of momentum? One has to be careful here, the momentum of
this volume is increasing, when the molecules come from above to below whereas, decreases
molecules goes from below to above. So this is going to be equal to j m y minus
minus j m y plus because what goes above is decreasing the momentum, what comes below
is increasing the momentum. And if I take this
Therefore, net flux of momentum is equal to j y m minus minus j y m plus. This is going
to be equal to nm u x at y plus 2 by 3 lamda minus nm times u x at y minus 2 by 3 lambda
times v rms times this undermined constant. Once again, I can do the same Taylor series
expansion. I assume that v rms is not changing too much because the temperature is not changing
too much and the density - number density and mass of the molecules is approximately
the same. So, this will give me a v rms nm into u x at y plus 2 by 3 lamda du x by dy
at y minus u x at y minus 2 by 3 lamda du x by dy.
This is equal to 4 by 3 a v rms nm lamda times du x by dy. So, this is the flux of momentum.
The flux of momentum is nothing, but the shear stress times tau xy. Newton’s law of viscosity
- tau xy is equal to mu du x by dy. So, clearly this coefficient mu is just everything that
is sitting here in front. The coefficient of viscosity is just the coefficient
that appears in front of the velocity gradient; it is this coefficient. Therefore, the viscosity
is approximately v rms nm times lamda. So, that is the viscosity of the molecular gas.
So, therefore, the viscosity is equal to some undetermined constant. Let us call that as
A nm v rms times lamda. The kinematic viscosity, we have discussed that earlier, when we write
the diffusion equation. The flux of a quantity per unit area per unit time is equal to diffusion
coefficient times the gradient of the density of that quantity. In this case, momentum;
momentum density is just rho times u x in the x direction - momentum per unit volume. Therefore, equation that
I had earlier is equal to mu into du x by dy. It can also be written as mu by row into
d by dy of rho u x, momentum density, where mu by row is equal to nu is the kinematic
viscosity. So, I have an expression for mu here. So, mu by row, row is the mass density.
In the expression for viscosity here, n is the number of molecules per volume times m
is the mass of the molecules. So, n times m is just rho, the mass density.
So, therefore, the kinematic viscosity is just some constant v rms times lamda. Kinematic
viscosity is momentum diffusivity. Diffusivity of all mass, momentum and energy have to have
dimension of length square by unit time. Clearly this has the dimension of length square by
unit time, but it is more than that. In the case of mass diffusion, we found out that
the mass diffusion coefficient was some constant times v rms times lamda; momentum diffusion
constant is the same in this case, some constant v rms times lamda.
So, both the mass and momentum diffusion are approximately of the same magnitude in gases.
That is no surprise. It is because the diffusion process for both mass and momentum are due
to the physical motion of molecules in a gas. So, in both cases we will get the momentum
diffusion constant to be roughly of the same magnitude. The viscosity in the gas, I got
it up to an undetermined constant here in a manner similar to diffusivity, I can do
a more detailed calculation and that detailed calculation will give me the constants in
the expression for the momentum diffusivity of the gas.
You can see here that viscosity is equal to A nm v rms times lamda and if I write out
the expressions for v rms 3 k T by m and this is 1 over root 2 pi nd square. So, you can
easily see that the number density at the bottom and the top cancel out and the viscosity
is independent of density in gases. If we do a more detailed calculation, we will
get the viscosity is 5 by 16 d square into mkT by pi power half. So, this is what we
get by more detailed calculation. We can see the dependencies are all correct. It has mkT
power half and because there is mass here, one over mass here and this is 1 over d square.
If I actually calculate the value of this, we will get L inverse T inverse, as we got
from dimensional analysis. So, for a gas, the viscosity increases with
temperature proportional to T power half, for a dilute gas. The viscosity is independent
of the density for a dilute gas. The reason it is independent of the density because the
reason is momentum transport is proportional to the number of particles, but, the mean
free path goes 1 over the number of molecules per unit volume; that cancels out and for
that reason, it is largely independent on the number of molecules.
Next, we will go kinematic viscosity and the momentum diffusivity in liquids. In liquids,
it turns out, when we discussed mass diffusivity earlier I told you the mass diffusivity in
liquids much smaller than the mass diffusion in gases. The reason is because the diffusion
of the mass requires the physical motion of molecules in liquid and that requires cooperative
motion because for one molecule to travel, other molecules have to move out of the way.
Momentum diffusion does not require the physical motion of molecules and for that reason the
momentum diffusion can happen much larger and much faster in liquids than mass diffusion.
The excess momentum can be transmitted from one molecule to the other due to molecular
interactions. It does not require the physical motion of molecules and for that reason, the
momentum diffusion in liquids is much faster than mass diffusion liquids.
We look at the details in the next lecture. I will go through these arguments for diffusion
of mass and momentum once again, complete the arguments for diffusion of energy in the
next lecture and then we start looking simple configurations.
So, basically in this lecture what I have tried to convey to you is the physical understanding
of diffusion and why it takes place. Diffusion takes place because you have changes in concentration
across the surfaces. If I have more molecules below and less molecules above, there is no
net flow, but because of the difference in concentration there is going to be a net motion
of molecules. That is true, whether it is mass, momentum or energy. In the case of gases,
the diffusion mechanisms are the same for all cases and we saw already the mass diffusivity
and momentum diffusivity are approximately the same.
We will see later that energy diffusivity is also approximately the same. In liquids,
they can be very different because the mechanisms are different. We will continue this discussion
in the next class, will see you next time.