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In today's lecture we shall take consider further problems in which, we shall apply
Reynolds flow model for predicting mass transfer rates; and the new problems are condensation,
transpiration cooling, volatile fuel burning, drying, and dissolution of solid in a liquid.
These are the vast range of problems in which Reynolds flow model can be applied, and I
have selected a few, so as to help you understand how easily one can estimate the mass transfer
rates, and arrive at very crucial engineering decisions in such problems. So, let us take
the first problem of condensation.
So, the problem statement is as follows; consider condensation of steam at one atmosphere on
the outside of a copper tube, so here is a copper tube, this is the symmetry line of
the tube of radius r I, and the wall thickness is r o minus r i through which cooling water
is flowing and the steam is on the outside of the tube; the tube is made of copper, its
diameter is 2.5 centimeter i d, and 2.9 centimeter o d, the tube carries cooling water at fifty
degree centigrade. Calculate the steam condensation rate on this
tube when steam is pure and saturated, and b steam is mixed with twenty percent air by
mass; when steam is mixed with air as you know the condensation rate reduces, and we
was to estimate how much is the reduction. Further data given are assumed that condensate
film thickness on the tube is .125 millimeters; the conductivity of copper is 300 watts per
meter kelvin that of water is 0.68 watts per meter kelvin.
The heat transfer coefficient on the tube side h cof i is 4620 watts per meter square
kelvin, we are going to assume t ref equal to t w where lambda ref the latent heat is
2257 kilo joules per kilo gram. Now, you are also given another information that if there
was single phase flow of air in this case, then the h coefficient without mass transfer
would be 115 watts per meter square kelvin. So, this is the value we shall use to estimate
g star; now, remember you can obviously solve this problem from heat transfer theory, and
you are always used heat transfer theory for solving problems of this kind, but we are
going to treat this as a mass transfer problem; so, my first task would be to show equivalence
between Reynolds flow model, and the heat transfer theory of condensation and that is
what i intended to do on the next slide.
So, in our case b would be omega v infinity minus omega v w divided by omega v w minus
1 and that would equal h m infinity minus h m w h m w minus h t l plus q l by n w and
that would equal n w by g. Now, as I said we are going to take t ref equal to t w, then
h t l will be of course 0, and h m w will be lambda ref into omega v w, there will be
no sensible heat contribution to h m w, because t ref is equal to t w, but h m infinity would
equate to c p m that is the mixture specific heat multiplied by t infinity minus t w plus
lambda ref into omega v infinity; so, if we substitute for h m infinity and h m w in this
expression and also make use of the definition of b and n w equal to g b.
Then substitution would show that n w can be written as q l, that is the conduction
heat transferred inside the transferred substance that is in the liquid film, divided by c p
m into t infinity minus t w divided by b minus lambda ref equal to g into b.
So this is our mass transfer or Reynolds flow model formula for calculating condensation
rate; but from heat transfer theory we know that q l is actually written as h condensation
into t w minus t s, where t s is the outside surface temperature of the tube.
And for pure steam omega v infinity will be equal to 1, so b will be equal to 1 minus
omega v w divided by omega v w minus 1, and therefore b will be equal to minus 1; and
hence our formula would simply read as n w equal to minus h condensation t w minus t
s c p m into t w minus t infinity because b is minus 1 plus lambda ref.
Now, this is precisely the formula you have used in heat transfer theory to calculate
condensation rate when steam is pure; so, thus our mass transfer formula accords with
the heat transfer formula with h condensation equal to condensation heat transfer coefficient,
and therefore there is a complete equivalence between the two.
So, let us turn to our part a of the problem when steam is pure. So, in our problem t s
is the outside tube wall temperature is not known, but the cooling water temperature t
c is known, and therefore we can invoke the notion of the total heat transfer coefficient
and write the condensation heat transfer Q l equal to divided by universal heat transfer
coefficient u into t w minus I mean total heat transfer coefficient u into t w minus
t c, where simply what I have done is 1 over u is the total resistance is equal to the
resistance due to inside heat transfer coefficient plus resistance due to thickness of the copper
wall and its conductivity. This is the resistance due to the thickness
of the liquid film having conductivity k liquid that is the water film; so, there are three
resistances adding up to 1 over u, and substitution if I substitute these values from the data
given here h cof i is 4620, k copper is 300, k water .68.
Then it works out that the total heat transfer coefficient u will be 2663 watts per meter
kelvin. Now, for pure steam at p equal to 1 atmosphere, t w t infinity will equal t
saturation, and that would equal hundred degree c.
And therefore, our formula is minus n w equal to g equal to h cof by c p of steam, and h
cof for natural convection was given as a 115 watts per meter square kelvin, and therefore
we readily calculate g is equal to 115 divided by 1.88 into 10 is to 3, equal to .06117 kg
per meter square also, so this is a straightforward application of our formula to estimate n w
which is negative n w equal to that; and from our model writing q l equal to u into t infinity
minus t c that is 2663 into 100 minus 50 into 1880 which is the specific heat of the mix
of the vapor into t w minus t infinity both are 100 degrees, and therefore, that term
makes no contribution divided by the lambda ref which is 2257 into 10 raise to 3, and
the result is .059 k g per meter square second. So, the two results are very very close .0617
and .059; now, of course, the difference arises mainly because we assume h cof divided by
c p v, and assuming h cof given is of the right order magnitude that the two results
are extremely close, and that verifies our Reynolds flow model for estimating condensation
rate.
Negative sign, of course, in both these results minus n w equal to that, and minus n w equal
to minus 059 indicates condensation. We now turn to the part b of the problem where there
is twenty percent air in the steam, and therefore omega v infinity will be .8, and q l as before
will be u into t w minus 50, but t w is not known, and therefore, our formula for n w
will be 2663 t w minus 50 divided by c p m into 100 minus t w divided by b minus lambda
ref and that would equal h cof o by c p m l n 1 plus b.
That is the formula for where b lambda ref and c p m all functions of t w, but we do
not know t w, and therefore we must do iterations; so, the thing is you assume t w, and therefore,
evaluate omega v w from saturation condition there and that gives us the value of b. Knowing
omega v w we calculate mean specific heat as 1614, and also evaluate lambda ref equal
to 2283.2 into 10 raise to 3, so the left hand side of this relationship gives us minus
.046, whereas the right hand side on here gives us minus .05 when t w was 90, so obviously
there is a difference in the between the left and right hand side, so we take next guess
of 91 degree centigrade.
And we find that the left hand side is now minus .0472, and the right hand side is minus
044, so obviously the result must be in between the two; and we now take 90.5, and we find
that the result is minus .0475 at the left hand side and we accept that result as nearly
correct, and take n w equal to minus .0473 k g per meter square second; now, you will
see that the our earlier result was minus .059, whereas now the result is minus .0473,
and therefore the condensation rate has reduced because of the 20 percent air in steam; and
this is a well-known problem in condensers, because the condenser operates at a very low
pressure, there is always a chance of air being ingressed, and therefore, the rate of
condensation reduces and that is why air ejectors are used in condensers.
This is just a problem of that variety which checks out what we normally do in order to
prevent falling of condensation rate, we always remove air from the steam which enters the
condenser. I will now take up the problem of transpiration cooling, which is always
used to protect surfaces which are exposed to very high temperature gases. So, consider
a problem as given here.
A porous metal surface is swept by air at 540 degree centigrade that is very high temperature.
Now, since metal oxidizes at 425 degree centigrade. It is decided to keep the surface temperature
down to 370, so our t w we want is at 370 by blowing gases through the pores. For this
purpose, three candidate gases available at 35 degree centigrade are considered, one is
air itself that is you inject air in air, you inject helium in air, and thirdly hydrogen
in air. Calculate the supply rate of each gas assuming operating g of 370 kg per meter
square hour; and we assume that the in all three cases of injection g mol as remains
constant. In case of air assuming constant specific
heat we assume that between 370 and 540, there is not much differences specific heat of air,
then the problem becomes not that it cannot be handled with variable specific heat, but
for simplicity c p is equal to t infinity minus t w c p t w minus t t, so that simply
becomes 540 minus 370, 370 minus 35 which is t t, and that is equal to .5074, that is
the b.
And therefore, n w of air would be g times b equal to 187.75 kg per meter square per
hour, so that is the answer for the part one of the problem. Now, we consider helium and
hydrogen - part b; so, in this case now specific heat of helium is 5.25 kilo joules per kg
kelvin and c p a infinity is 1.1 kilo joules kg per kelvin.
So therefore, if I take t ref equal to t w, then you will see I do not have to calculate
mixture specific heat, because it will simply mean t w minus t ref gets canceled here as
t w minus t ref gets canceled here, this is h infinity, and this is h t, and that would
simply be c p a into t infinity minus t ref divided minus c p helium into t t minus t
ref, so I will get 1.1 into 540 minus 370 divided by minus 5.25 into 35 minus 370 b
is .0163. So, b for helium is smaller than that for air, because of high specific heat
and you will see n w helium is g times b will be about 40 kg per meter square hour, so much
reduced quantity of helium is required compare to air.
Now, let us take the case of hydrogen. In this case hydrogen specific heat is 14.5,
and hydrogen is going to burn in such a hot environment; so, we assume a simple chemical
reaction hydrogen plus half o 2 equal to h 2 o giving r s t is equal to 16 by 2 equal
to 8, that is the stoichiometric ratio; for this case is 16 by 2 equal to 8, and the latent
heat of hydrogen is 118 mega joules per kilogram, and therefore taking h equal to c p m into
t minus t ref plus del h c by r s t omega o 2.
As we know we have associated the latent heat with oxygen, then therefore we divide this
by r s t; and if we take t ref equal to t w we have b equal to h infinity h w of course
will be 0, because omega o 2 cannot survive at the surface as well as t is equal to t
ref, and we will have minus c p h two into t t minus t w which will be the h t, this
is 1.1 c p a infinity is 1.1 into 540 minus 370 plus 100 and 18.4 into 10 raise to 3 divided
by 8 into .232 divided by minus 40.5 into 35 minus 70.
And now, the b increases to .745, and therefore n w of hydrogen will be g into b 275.8; remember,
we are using hydrogen which is going to burn to keep the surface cool at 370 degree centigrade.
And you will see now that, in the three application if we find that the amount of hydrogen required
will be greater than that for air and that for helium, this sort of simple calculations
enable us to select the right kind of gas to inject through the porous surface depending
on our requirement. I will now take up the next problem and that
is of missile cooling. Now, missiles as you know travel at very high altitudes at very
very high speeds of the order of 5 to 7000 meter meters per second very similar to the
kind of velocities that are encountered when the reentry vehicle enters the upper atmosphere
at around 10000 meters per second.
So, these are very very high velocity projectiles which have to be kept cool, because of the
viscous heating that takes place near the surface. So, the problem reads as follows,
consider axis symmetric stagnation point of a missile travelling at 5500 meters per second
through air where static temperature is almost say 0 k. It is desired to maintain the surface
temperature at about 1200 degree centigrade by transpiration cooling of hydrogen at 38
degree centigrade. Evaluate B and N w, given g star in this case equal to .467 kg per meter
square second. So, here the value of g star is again given;
now, because of the high velocity here we must account for the kinetic energy contribution,
and define h m equal to c p m t minus t ref plus delta h c divided by r s t omega o 2,
which is as usual, plus v squared by 2000 as the kinetic energy contribution to enthalpy
in kilo joules per k g, therefore, v squared by two into 1000 that is just what in kilo
joules per kg will be. So, taking t ref equal to t w so that h m
w again is 0, then the first x, the numerator will be h infinity divided by minus h t will
be c p h 2 into t t minus t w, and you get ah c p a infinity equal to 1.1 minus 0 minus
1473, because t infinity is almost taken as 0. The wall temperature is 1200 degree centigrade
or 1473 kelvin plus 118.4 into 10 raise to 3, which is the delta h c of hydrogen divided
by 8 as before into .232, which will the omega o 2, which is that which is the mass fraction
of oxygen in air plus 5500 square divided by 2000 divided by minus 14.5 into 38 minus
1200. So, this gives us the large v of value 1,
and n w will be g star into l n 1 plus b will give you g star is .467, and therefore .325
kg per meters square second. Missiles of this type have to carry a certain amount of gas
with them, so that in order to keep the surface cool, the particularly the stagnation point
of the missile has to be kept very very cool, because it is it is travelling at a very very
high velocity; under such conditions a hif, you know the time of flight knowing the mass
transfer rate as we have calculated here, we can calculate so knowing the time of flight
that is seconds and knowing the surface area over which you are going to inject the gas.
We can calculate the amount of hydrogen that the missile must carry with it; so, problems
of this kind give you a first estimate of how much hydrogen to carry in a missile without
solving any differential equation.
Let us now look at another problem, and that is of burning of a volatile fuel. So, in a
diesel engine liquid fuel c twelve h 26 with heat of combustion 44 mega joules per k g
and specific gravity 0.854, and latent heat 358, and boiling point 425 degree centigrade
is injected in the form of small droplets.After ignition delay, part of the fuel vaporizes
and burns abruptly and the remainder of the fuel burns as fast as the fuel vaporizes.
Estimate the burning time of a 5 micron droplet. You are given that temperature in the cylinder,
the gas temperature of the cylinder is about 800 degree centigrade, and the conductivity
of the fuel is .0463 watts per meter kelvin. Now, from stoichiometry for c 12 h 26 fuel
the oxygen to fuel ratio will be 18.5, which is m plus n that is 12 plus 26 by 4 will be
18.5 multiplied by 32, which is the molecular weight of oxygen divided by 170, which is
the molecular weight of the fuel, and therefore the oxygen to fuel ratio will be 3.482.
We define h m equal to c p m into t minus t ref plus del h c by r s t into omega o 2.
And assume the droplet to be at it is boiling point, so all temperatures t ref t b p t w
and t t are all at the boiling point of 425 degree centigrade. So, that h m w h t l and
q l, there will be no conduction heat transfer inside the liquid fuel; and therefore we will
simply have b equal to h m infinity minus 0, which is h m w h m w again 0 minus h t
l, which is 0, and this will be simply h f g, because q l by n w would be..., as you
can see q w will be n w h t w minus h t l will be simply n w into h f g, and therefore
our formula will simply be c p infinity into t infinity minus t b p plus del h c by r s
t omega o 2 infinity by h f g and taking c p infinity equal to 1.15 800 minus 425, which
is the boiling point 44 into 10 raise to 3, which is the heat of combustion into divided
by 3.482, which is r s t into .232 divided by 348, which is the value of h f g you have
been given there, and therefore 9.394 very very high b in case of liquid volatile burning.
Now, let us go on to calculate the burning time of a 5 micron droplet.
Since b is large, the instantaneous burning rate and the burning time would be given by,
m dot into gamma h by r w 4 pi r w square l n 1 plus b and that will give us t burn
equal to rho l into d w i square 8 into gamma h into l n 1 plus b. Now, at 800 degree centigrade
conductive heat of air is .075 such calculation the mean conductivity is evaluated by as .4
times fuel conductivity plus point six multiplied by air conductivity, and therefore the mean
conductivity will be .06353. And mean specific heat we can take as 1.2 kilo joules per kg
kelvin, and therefore gamma h which is k m by c p m would be .06353 divided by twelve
hundred equal to 5.29 into 10 raise to minus 5 kilograms per meter second.
Now, you are also given rho l, the liquid the specific gravity is .84, and therefore
rho l will be 853 kg per meter cube, and you will recall that our..., and if we substitute
that here t burn equal to 853 5 into 10 raise to minus 6 raise to 2 into 8 into 5.29 into
10 raise to minus 5 l n into 1 plus 9.394, which is the value of b, and that gives us
0.0215 milliseconds. Now, of course, here I have estimated the
burning time based on assuming that the environment inside the diesel engine is more less stagnant,
that is at the top deck center where the burning takes place there is not much velocity; in
reality there would be some velocity between the injected droplet and the gas which is
turned as a result of the piston movement. And therefore, there would be some enhancement
in the effective value of gamma h, because of convection pass the droplet; of course,
the his role will be very very small, because the droplet diameter is very very small of
5 microns, and therefore the answer is reasonable. Such time is very very important, because
this time affects the cutoff ratio of the diesel engine, and therefore its efficiency.
As you know we want very small cutoff ratio, and if all the fuel burns out in a smaller
time then the cutoff ratio will be smaller, and therefore, it is of crucial importance
to estimate what the burning time of a droplet would be.
Of course, in a diesel engine there will cloud of droplets and in our estimate of b; we have
assumed that each droplet experiences omega o 2 in infinity equal to .32. In reality some
droplets that is at the front of the cloud will experience omega o 2 infinity equal to
.232, but those that follow will be facing a mixture of burnt products and air, and therefore,
they will experience somewhat lower value of omega o 2 infinity, and therefore the b
will be smaller, they will take a little longer then to burn.
But nonetheless the estimate that we have provided here 0.215 millisecond is a very
valuable one in accessing how quickly the fuel shall burn inside a diesel engine.
We now take up another problem and that is of convective drying of cloth in a laundry
dryer. So, the problem statement is as follows, in a laundry dryer dry air is available at
1 bar and 20 degree centigrade. Now, of course, this is a cold air, and therefore the engineer
wants to try out mixing of this air with super-heated steam at 1 bar and 250 degree centigrade.
Now, use of steam in for super-heated steam for drying is quite routinely used in textile
industry particularly in a device called the stenter, where the wet printed cloth is to
be dried at a very fast rate; instead of using air which gives you rather, because air has
to be heated somewhere, and that air has a lowest specific heat than specific heat of
the of steam, and therefore what is preferred is super-heated steam drying rather than air
drying. So, we want to examine here the possibility
of mixing 1 bar 250 degree centigrade super-heated steam with air which is available at 1 bar
and 20 degree c. And let us see if the effective mixtures which will flow pass the drying cloth
would actually dry it fast. So, examine the effect of mixing in the range of 0 omega v
infinity to .5 assume g is unchanged with change in omega v infinity.
Now, for superheated steam at 1 bar and 250 degree centigrade from steam tables we have
we can get h v infinity equal to 2974.3 kilo joules per k g. There will be adiabatic conditions
at the drying surface, and therefore q l equal to 0, and hence our formula would be b omega
v infinity minus omega v w omega v w minus 1 equal to omega h m infinity minus h m w
h m w minus h t l, where omega h m infinity will be the enthalpy of air plus enthalpy
of steam weighted by the mass fractions, 1 minus omega v infinity is the mass fraction
of air, and omega v infinity would be the mass fraction of vapor or the steam.
At the surface, we do not know the value of t w, but we can take this as 1.005 t w 2503,
because t ref is taken as 0, omega v w, and h t l would be 4.187 into t w, where t w and
omega v w are related by equilibrium relation given in lecture thirty-seven, and therefore
we must iterate as usual to assume t w, therefore calculate v w, therefore calculate h m w,
and see whether the left hand side agrees with the right hand side.
So, we need iterative solution, and on the next slide I will show you the iterative solutions
for a range of values of omega v infinity varying from 0 to .5; so, if omega v infinity
is 0, that is only pure air is supplied at 20 degree centigrade, then t w actually turns
out to be 6 degrees only, and b is .0056, therefore we expect very very low rates of
drying. At .01 that is 1 percent steam addition when
would have thought that the drying rate will actually increase, but you will see that the
drying rate has actually decreased to .00274 b is decreased. If I make it .03, that is
three percent, then it still further decreases to .00018, but the temperature increases of
the surface to 32 degree centigrade; now, at .04 and .05, you one finds that the b values
are extremely small, 10 raise to minus 5, 10 raise to minus 6 of that order.
And therefore, it is extremely difficult to balance b m equal to b h, and find t w exactly,
so these are skipped now, and I go to eight percent air, and I find that it is .009, so
it is better than three percent air, but still less than that with pure air. If I go to 10
percent air however I get .00225 again the drying rate will increase, but still lower
than the value corresponding to pure air. But at 20 percent I get .0122 that means,
b has now doubled, and 20 percent steam mixing would achieve results in the desirable direction
at 30 percent, of course, still further improvement in b. At 40 percent still further improvement.
And at 50 percent .0517 to still further; so, drying rate is a non-linear at small mass
fractions omega v infinity when they are small, it first decreases with increase in omega
almost goes to 0, then starts increasing again and beyond .1 it exceeds the value of pure
air case very interesting result. So, here is a comment that it is difficult
to balance b m equal to b h exactly when omega v infinity is around .05, because b itself
tends to 0; so, compared to dry air, drying rate improves monotonically beyond omega v
infinity .02, and of course, if you went for a pure steam you will certainly get much higher
drying rate compared to pure air which is omega v infinity equal to 0.
So, such industrial irrelevant problems also can be very effectively solved by Reynolds
flow model. Finally, I take up the case of solid dissolving in liquid, and I am taking
up the case of a thin plate, now there is no gas involved here, nor is there any chemical
reaction, it is a dissolution of solid into a liquid; in fact it is a salt slab as you
can see here in the problem statement, a thin plate 15 centimeters by 15 centimeters of
solid salt is to be dragged through sea water edge wise at 20 degree centigrade with a velocity
of 5 meters per second sea water has salt concentration of 3 percent by weight.
Saturated salt solution in water has concentration of 30 grams per 100 grams of water at 20 degree
centigrade; assuming transition criterion of fraser and milne determine if transition
will occur in this particular case and estimate the rate at which salt will go into the solution
from the solid plate. Take for the salt solution Schmidt number is 745 which is very high the
nu kinematic viscosity of water's sea water is 10 raise to minus 6 meters square per second,
and the salt specific gravity is 2.163. So, first part of the question we want to
examine whether the when the sea water flows over the plate there will be transition or
not now this is a case of m equal to 0, that is the pressure gradient is 0, and delta two
star will be .645 Reynolds x to the minus .5. And from faser and milne, for m equal
to 0 r e delta two will be 163 plus exponential of 6.91or 1165.2 will be the r e based on
delta 2, and therefore if I substitute that here this result and this result.
Then we would get r e x transition equal to 3.08 into 10 raise to 6 or u infinity equal
to 5 meters per second, which gives x transition at 61.6 centimeters per second, which is of
course greater than our plate which is 15 centimeter. And therefore, over the plate
transition will not occur we will simply have laminar boundary layer. Now, let us go on
to the part b of the problem, where we wish to estimate the rate at which salt will go
into the solution.
So in this problem omega infinity is .03, because that is the sea water concentration
you have been given that; at the surface of the salt it will be 30 grams per hundred grams,
and therefore 36 divided 100 grams of water, and therefore it will be 36 divided by 136
equal to .2647, omega t which is pure salt will be 1, and therefore b will be .03 minus
0.2647 divided by .2647 minus 1 equal to 0.3192. And therefore, the plate Reynolds number of
course here is 5 meters per second into the length of the plate, which is 0.15 divided
by kinematic viscosity which is 10 raise to minus 6 is 7.5 into 10 raise to 5. And therefore,
Sherwood number would be g star l divided by rho m into diffusivity will be equal to
.664 r e l to the .5, Schmidt number equal to .33will be 5.99.3 that would be the Sherwood
number. Now, in order to calculate rho m, we take
rho m equal to rho water plus 1 minus omega mean which is the mass fraction of water plus
omega mean into rho of salt, where omega mean will be taken as .03 in the sea water, and
.2647 at the surface, as we have shown here, and that is mean is .147, so therefore, mean
density would be 1169.2, and therefore g star taking l equal to .15, rho m equal to 1169.2,
and diffusivity value has been given to you, because Schmidt number is known, and nu is
known. And therefore, we can get the value of diffusivity,
therefore we get g star is estimated at 192 kilograms per meter square hour; and therefore,
n w will be g star into l n plus into 1 plus b equal to 53.2 meter kilograms per meter
square hour; and therefore, the mass loss from two sides of the plate will be 53.2 into
two into .15 square, which is the area of the plate and the two sides.
So, 2.394 kg per hour would be the answer for the rate at which the salt will go into
the solution. So, this is a departure from other problems of gaseous type, we here we
have mass transfer into liquid from a solid, so with this I complete all our discussion
on essentially force convection heat and mass transfer. In the first 10 lectures I dealt
with laminar flow, the first 10 lectures I develop the main theory of laminar flows,
external boundary layers then took up the case of internal laminar internal flows.
Simple as well as complex shaped, but all under force convection situation then we spent
10 lectures that is lecture number twenty-one to lecture number thirty on turbulent flows,
where we discussed the formal and predictive aspects of turbulent flows.
And develop methods for estimating calculating external boundary layers both starting from
laminar through transition to turbulence over external surfaces; as well as we applied phenomenology
as well as analogy methods to determine nusselt numbers for internal flows like pipe flows
and included effects of roughness and so on so forth. So and then from lecture number
thirty onwards we turn to convective mass transfer, where we develop the Stefan flow
model, the Couette flow model, and the Reynolds flow model as proxies for the boundary layer
flow model.
And in the last few lectures I applied these models to practical problems of engineering
relevance. So, our conclusion then is that, the that the algebraic Reynolds flow model
with property corrections is a good proxy for the boundary layer flow model, because
mass transfer coefficient is evaluated from h cof infinity at v w equal to 0 for the corresponding
heat transfer situation, and we have number of co relations for h cof v w equal to 0,
which we can readily use for the corresponding mass transfer situation.
And therefore, estimate g star and go on to calculate the mass transfer rate, it is this
feature which obviates the need for solving complete set of boundary layer equation; the
one d Stefan flow model provides a reliable solutions in diffusion mass transfer, the
one d Couette flow model though very approximate provides mean for estimating effect of property
variations in a boundary layer. And we showed that the Couette flow model
actually does predict separation of variable property mass transfer rates when property
correction is applied as this found from the Reynolds flow model or invent from the boundary
layer flow model. So, with this really I complete all issues of convective mass transfer at
the interface as well as convective heat transfer, but all under forced convection situation.
And in the remaining two lectures I am going to take up two special topics, and these are
described here at the bottom of the slide; so, we will consider two special topics - one
is of natural convection boundary layers in which I will consider both heat transfer as
well as mass transfer. And secondly where heat and mass transfer again play an important
role is laminar diffusion flame, that is it will be heat and mass transfer with chemical
reaction, but in a free jet situation, there is no wall, but we the what is of importance
is to predict the length of the flame and the width of the flame. And these are two
special topics which I think are relevant to convective heat transfer, and therefore,
in the last two lectures I will be dealing with these two topics.