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Today we will begin with the topic of condensation and boiling; we will be studying the heat
transfer during this change of phase processes. Now, first of all, the obvious question is
where are we interested in, what situations are we interested in studying heat transfer
during condensation and boiling. Well, if you look, say at a power plant - any thermal
power plant - you got change of phase occurring when water is converted into steam in a boiler
or sometimes in an evaporator and you got change of phase occurring when the steam which
has been expanded through a turbine is condensed in a steam condenser. Or you, if you look,
say at a refrigeration system - vapor compression refrigeration system - there also we have
change of phase occurring when the refrigerant is condensed in a condenser after having given
up its heat in the evaporator. So, there are numerous such situation; I have just mentioned
2 where we are interested in knowing the heat transfer rates during change of phase that
is either during condensation or during boiling. So, our purpose in this topic when we are
studying this topic is to derive expressions for the heat transfer coefficient in some
situations for some geometries and with the help of these expression or with the help
of these correlations, one is able to design the equipment in which either condensation
or boiling is taking place.
Now, first a general comment; a general comment which is worth noting is the values of h,
the values of the heat transfer coefficient, during condensation and boiling are invariably
high as compared to the values we are used to in single phase heat transfer. For example,
just to illustrate the idea, suppose I have got single phase heat transfer natural convection
to a gas. A typical value of h is 5 or 10; h is typically 5 or 10 Watts per meter squared
Kelvin. Suppose I have got heat transfer occurring and natural convection to a liquid; typical
value of h is00 200 Watts per meter squared Kelvin. On the other hand, suppose I have
got forced convection; depending on the velocity the value of h forced convection to a gas
maybe 20, 30, 40, may go up to a 100, 150 Watts per meter square Kelvin, at very high
velocities maybe 200 Watts per meter squared Kelvin. And similarly, if I have got single
phase to a liquid, typical values of h in forced convection range from a few00 that
is 200, 300, 400 at low velocities to maybe a few000 - 3000, 4000 - at reasonably high
velocities used in industrial practice.
In contrast to this, in condensation and boiling in most of the types of condensation and boiling
that takes place, values of h are invariably in thousands - 5000, 10000, 20000 Watts per
meter squared Kelvin are typical values of h during condensation and boiling. There are
some exceptions; I don't want to say it is always the case. There are some exceptions
which you will point out when we go further but in general keep that in mind, values of
h during a change of phase process that is during condensation and boiling are invariably
in thousands. 1000, 2000 are low values; typically 5000, 10000, 20000 are values that we encounter
all the time so, that is just one general comment.
Now, let us let us first of all study condensation; we are first of all going to study, take up
the study of condensation. When does condensation occur? That is the first statement we want
to make, that is the first the thing we want to clarify.
When does condensation occur on a surface, when? Let us say I have some surface, let
us say it is a vertical surface and it is maintained at a temperature Tw with the help of some fluid flowing on this side,
the back side. And on the right hand side I have a vapor, some vapor - it maybe steam
or it may be some other vapor; it doesn't matter. And let us say the saturation temperature
of that vapor corresponding to the vapor pressure is Ts - it is a vapor which is at some pressure
Ps and the saturation temperature corresponding to it is Ts. It is a saturated vapor, the
pressure Ps and the temperature Ts.
Now if Ts is greater than Tw, that is the wall is maintained at a temperature lower
than the saturation temperature of the vapor which is in contact with it, then this vapor
is going to condense on this surface. So, condensation occurs when a vapor with a saturation
temperature Ts comes in to contact with a solid surface which is maintained at a temperature
Tw; Tw being less than Ts, that is the, that is when condensation occurs. Now, the nature
of the condensation which occurs, the nature of the condensation which occurs or the type
of condensation which occurs is of 2 types. It maybe of what we call as the film type
which we call as film condensation or it may be in the form of droplets which we call as
drop-wise or drop condensation.
So, the condensation which occurs on the surface when this condition Ts is greater Tw is satisfied
maybe of 2 types - film condensation or drop condensation; film condensation occurs, film
condensation occurs when? Occurs when the liquid condensate, when the liquid condensate wets the surface. The surface tension between
the liquid and the solid is such that the liquid condensate wets the surface when that
happens film condensation occurs. And the word film is descriptive enough; it means
that the vapor on condensing forms some kind of film on that solid surface; that is why
we call it film condensation. On the other hand, drop condensation or drop-wise condensation
occurs when, let me write it on another page.
Drop condensation occurs when the liquid does not wet the surface. When the liquid formed
during condensation does not wet the surface then it accumulates in the form of drops on
that solid surface and we get drop-wise condensation. So, when the liquid does not wet the surface
the condensate collects in the form of droplets. These droplets grow in size as more and more
condensate forms on them which grow in size. Sometimes, they join up with each other; if
they are near enough, coalesce with neighboring droplets
and eventually when they become large enough, these droplets roll off the surface. This
is the process of drop-wise condensation.
The requirement is the liquid condensate should not wet the surface; when it doesn't it form
as droplets on the surface then the condensate forms, these droplets grow in size. If they
are near enough to each other some of them join each other and coalesce and then when
they become large enough. They cannot under the force of gravity, they may roll off the
surface and drop off completely. So, these are 2 types of condensation which occur, which
we observe from our daily needs also we can observe these 2 processes.
Now the interesting thing from a heat transfer standpoint is the rate of heat transfer during
the 2 types is quite different, that is what we need to note. The rate of heat transfer
during the 2 types of condensation
is quite different, that is what we need to note.
Let us say I plot, let us say I plot a graph showing and on in this graph on the x axis I have Ts minus Tw. Ts is the saturation
temperature of the vapor which is condensing on the solid surface and Tw is the wall temperature
which is slightly less than Ts. Let us say on the x axis, I plot Ts minus Tw and on the
y axis I plot the heat flux transferred during the condensation process, the heat flux in
Watts per meter square, typically if you collect experimental data in one case for film condensation
occurring on the surface, in the other case for drop condensation occurring on the surface.
And through the experimental data you draw some curves typically in drop-wise condensation;
you might get a curve something like this in film condensation, you might get a curve
something like this. This would be film condensation and this would be drop condensation, these
would be typical curves which you would get if you were to draw them through experimental
data.
Let us say we will get experimental data like this; we will get experimental data in condensation
like this and draw a smooth curve through them. You will get curve something like this.
What does this mean? What these graphs, what these graphs are showing is for the same temperature
difference; let us say for a particular temperature difference, some value - the heat flux during
drop-wise condensation - would be so, much and for the same temperature difference the
heat flux during film condensation would be so, much . If I go up for a particular value
of temperature difference during film condensation I will get so, much heat flux and during drop-wise
condensation I will get so, much heat flux. So, during drop-wise condensation one gets
much higher heat fluxes for the same temperature difference. So, obviously from a heat transfer
standpoint I would be interested in having drop-wise condensation.
That is the first one can see from this experimental data, that drop-wise condensation is desirable
because it gives us higher heat fluxes for the same temperature difference. But what
we find during practice is that in reality, suppose we have some condenser - maybe a steam
condenser or may be a refrigerant condenser - and we start condensing either the refrigerant
or steam or some other vapor on the surface which is being cooled. Typically, what we
find is that while one may get drop-wise condensation for a few hours, it is difficult to maintain
drop-wise condensation, does not continue to occur if you have the equipment running
for hours and hours - hundreds of hundreds of hours of operation.
So, while it is desirable to have drop-wise condensation, it is difficult to maintain
drop-wise condensation until one does some special things with the surface on which the
condensation is occurring. For typical manufactured surfaces on which condensation takes place
drop-wise condensation may occur initially but is difficult to maintain. Therefore, although
from a heat transfer standpoint one would prefer to have drop-wise condensation, one
designs heat transfer equipment during condensation under the assumption that the condensation
is of the film type. So, keep that in mind that we would, drop-wise condensation may
look desirable but we design heat transfer equipment under the assumption that the condensation
is of the film type so, let me just make a few notes here.
Drop-wise condensation is desirable, that is the point I made. Drop-wise condensation,
as we can see from the data shown in this figure but is difficult to maintain, difficult to
maintain; I am repeating what I said - difficult to maintain. Therefore, the usual practice
is to design heat transfer equipment; the usual practice is to design heat transfer
equipment under the assumption that condensation is of the film type - that is what we do.
Although it is desirable to have drop-wise condensation, it is difficult to maintain
and therefore the usual practice is to design heat transfer equipment under the assumption
that the condensation is film condensation. Now - mind you - even in film condensation,
although the heat transfer coefficients are less than in drop-wise condensation, they
are are still high. It is not as if they are very low; they are high but they are low compared
what we get in drop-wise condensation.
So, from now onwards let us focus our attention on film condensation because as I said heat
transfer equipment is designed under that assumption. We are going to first of all look
at the classical problem of film condensation on a vertical plate. It is the problem which
I sketched a moment ago and we will talk about that now in some detail.
We are going to, let me read out here, we are going to derive an expression for the
heat transfer coefficient for h during laminar film condensation of a pure stationary and
saturated vapor at a temperature Ts on an isothermal vertical plate at a temperature
Tw, that is what we are going to do. This is the classical problem which was, this problem
was first solved by Nusselt in Germany in 1916; it is the classical problem of condensation
and we are going to derive the expression which was derived then so, many years ago.
Let me again repeat that. Derive an expression for h - the heat transfer coefficient - during
laminar film condensation of a pure stationary and saturated vapor at a temperature Ts which
is condensing on an isothermal vertical surface at a temperature Tw.
Let us just look at the geometry a little for your, for the moment focus your attention
on the left hand side not on the whole figure which I am displaying or rather let us cover
the rest of the figure so, that your attention is not diverted. This is a plate, a vertical
plate here, and just to show the plate let us draw a double line here. This is a plate
like this; this is a plate, a vertical plate, and it has, it is being maintained at a temperature
Tw, maintained at a temperature Tw, a vertical plate being maintained at a temperature Tw.
And on the right side here, on the right side coming in contact with it is a pure stationary
saturated vapor at a pressure Ps and a corresponding temperature Ts and of course Ts is greater
than Tw .
So the vapor condenses on this plate because Ts is greater than Tw and it condenses in
the form of a film. We will assume that film condensation is taking place and the film
thickness here is shown, exaggerated actually. It is a very thin film that forms on the surface
and flows down but I am showing it here purposely exaggerated because we want to understand
the fluid mechanics of that film, its thickness, etcetera. So, in order to study it, we are
showing the thick of the film delta as a very exaggerated film thickness here.
The film, obviously condensation begins from the top so, the film will start with zero
thickness at the top and will grow in thickness as it as we go downwards. As I say, it is
the delta, the thickness of the film is exaggerated in the sketch. The coordinate system we will
adopt is y, z; y is direction at right angles to this vertical plate, z is the direction
along the plate and the origin is right here at the top. So, z is measured from the top
of the plate positive downwards, z is measured from the top of the plate positive in the
downward direction and the direction at right angles to the paper, the direction at right
angles to the paper, it will be obviously the direction x. And let us assume that the
plate is of width B, B at right angles to the paper - some constant value B. So, it
is a plate of width B on which condensation is taking place and we want to analyze the
process of this condensation.
As I said a moment ago, this problem was first analyzed by Nusselt and he analyzed it by
making a number of simplifying assumptions and basically applying the fundamental laws.
What are the fundamental laws? The fundamental laws are conservation of mass that is the
equation of continuity, the Newton's second law of motion and the first law of thermodynamics.
So, that is all that, I mean, Nusselt applied to this problem with certain simplifying assumptions
in order to obtain the value of h for this particular situation. Now, we are going to
go through that process of derivation. The assumptions which we are going to make are
the following, the assumption which we are going to make are the following. I will first
list them one by one and then of course as we go along I will justify why they are being
made.
First of all the assumption we are going to make is - the fluid properties are constant.
During the analysis, we will need properties like the conductivity of the fluid, the density
of the fluid, the viscosity of the fluid, etcetera. We assume some constant value for
these properties. If I go back to the sketch, the temperature of the liquid condensate is
going to vary a little inside. Here it will be Ts at this point, Tw at this point. Therefore,
if temperature varies, properties are going to vary a little but this variation for properties
will be exceedingly small and therefore it is a good assumption to say that we can take
some constant value of the properties. It is a very, an extremely good assumption to
make for this particular problem.
The second assumption we are going to make is, the liquid vapor interface is at the saturation
temperature Ts, liquid vapor interface is at the saturation temperature Ts. What do
we mean by this? Let me again go back to the sketch. This is the condensate forming this
curve line, this is the condensate film surface so, this is the interface right here. This,
at this interface the temperature is the saturation temperature Ts, all along this liquid vapor
interface the temperature is Ts which is also a good assumption to me.
The third assumption is that the liquid momentum effects in the liquid film are negligible.
Now, what do we mean by this? What we mean is the following - momentum effects in the
liquid film are negligible. Here is the liquid film; it grows in thickness and the liquid
is flowing downwards like this right in this film, because of the action of the gravity
it is flowing downwards along this vertical plane. When it flows downwards, it has a velocity
and if something has a velocity it has momentum. What we are saying is when you apply Newton's
second law of motion to a control volume there are terms in that equation which say that
the net forces in a particular direction are equal to the momentum change momentum flowing
out of the control volume minus the momentum flowing into the control volume. Let me repeat
that. Newton's second law as applied to a control volume says net forces in a particular
direction are equal to the momentum flowing out of the control volume minus momentum flowing
into the control volume.
What we are saying for this particular problem is that the momentum effects - the change
of momentum flowing out minus momentum flowing in - is negligible. In any control volume
that I have selected inside this liquid film momentum effects are negligible. Therefore,
in effect, what we are saying is the net forces acting on a the surface or the net body forces
and surface forces acting on any control volume are equal to zero. Momentum effects need not
be taken into account; we will come back to this when we take the particular control volume
in a moment. so, we will neglect momentum effects inside of the liquid in the liquid
film.
Assumption 4 - the vapor exerts no shear stress, vapor exerts no shear stress at the liquid
vapor interface. What we mean again by this is the following. As this liquid flows down,
it has a velocity and that velocity is relative to the stationary vapor which is next to it.
So, obviously, if something is moving, the liquid is moving and the vapor is stationary,
one thing must exert a shear stress on the other. So, there has to be a shear stress
at this liquid vapor interface because there is relative motion between the liquid and
the vapor in the vertical direction. What we are saying is, this shear stress is negligible
again, probably an exceedingly good assumption to make for the physical situation that we
are talking about.
Then the next assumption which Nusselt made and which we are making is the temperature
distribution in the film is linear. The film is very thin, the liquid film as I said a
moment ago, which is condensate film formed is a very thin film and Nusselt said let us
assume that the temperature variation in this liquid film is linear, it is a straight line.
So, we will make that same assumption that within the liquid film the drop from Ts to
Tw over the thickness of the liquid film, which is delta, will be a straight line. Again
I have probably an exceedingly good assumption to make.
And final assumption which Nusselt made was, he said enthalpy changes associated with sub-cooling
of the liquid are negligible. What he meant was the vapor first condenses at the liquid
vapor interface, vapor first condenses at the liquid vapor interface where the temperature
is Ts but the liquid inside, its temperature drops from Ts to Tw inside the film and Tw
is less than Ts so, that is the sub-cooling that is taking place in the liquid condensing.
Nusselt said these enthalpy changes associated with sub-cooling from Ts to Tw are negligible
compared to the latent heat lambda that we will have to take account of during the change
for phase and that probably is also an exceedingly good assumption to make.
So, these were the 6 assumptions that Nusselt made - fluid properties are constant, number
2 -liquid vapor interface is the saturation temperature Ts, momentum effects in the liquid
film are negligible, the vapor exerts no shear stress at the liquid vapor interface, the
temperature distribution in the film liquid film is linear and finally enthalpy changes
associated with sub-cooling of the liquid are negligible. These were the assumptions
which Nusselt made. Now, straightway you can see that if I look at assumption number 2
which says that the liquid vapor interface is at the temperature Ts and if I look at
assumption number 5 which says the temperature distribution in the liquid film is linear
and if I combine these 2, then I can see that the temperature profile in the liquid film
is going to be like this. In the vapor space, the temperature is going
to be constant Ts and then at the wall it is Tw, so, across the liquid film I am going
to get a straight line like this. So, the temperature distribution in the liquid film
and in the vapor space based on assumptions 2 and 5 will be as shown. As I am showing
you just now, that is where we have used assumption 2 and assumption 5.
Now let us move on; let us apply our fundamental laws. Now let us consider, let us consider
this is our liquid film. I go back to my sketch, this is my liquid film. Let us say I consider
inside the liquid film a differential control volume dy dz which I am showing just now,
shading. That means in the y direction at right angles to the plate inside the liquid
film, we take a differential dy and in the z direction that is moving downwards we take
a distance dz. In the x direction of course, we have a width B. It is a 2 dimensional problem
so, we have a width B in the x direction. so, take a differential control volume dy
dz inside the liquid condensate film.
Let us apply Newton's second law to this control volume. What does Newton's second law state?
Newton's second law will state - if I want to apply it for the z in direction - the direction
of flow - it will, say net z direction forces acting on this control volume on the liquid
inside this control volume, net z direction forces acting on the liquid. This control
volume are equal to the rate at which z direction momentum leaves this control volume minus
the rate at z direction momentum enters this control volume. That is what Newton's second
will state if applied to this control volume.
We have already made an assumption; what are that assumptions? The assumption was that,
let me repeat it and show it here. The assumption was, number 3 assumption was, momentum effects
in the liquid film are negligible. That means we are saying we will neglect the difference
between the momentum flowing out and the momentum flowing into this differential control volume
dy dz. So, what are we left with? Therefore, if we do make that assumption, we will now
get, let me write that down.
Now, what will Newton's second law as applied to this control volume for the control volume
dy dz? Newton's second law states net z direction forces, net z direction forces exerted on
the liquid on the condensate liquid inside the control volume equal to momentum z direction
momentum leaving z direction, momentum leaving the control volume minus z direction momentum
entering the control volume, that is Newton's second law as applied to the control volume.
In this particular case the terms on the right hand side of this equation we are saying are,
the difference between them is negligible, momentum effects are negligible. Therefore,
we are saying net z direction forces exerted on the liquid inside the control volume is
equal to zero. Now, what are the net z direction forces exerted on the liquid are of 2 types
- surface forces and body force. What are the surface forces in this case? Let me say
the z direction forces are surface forces and body forces. Now, let us look at the control
volume again. Here is my control volume dy dz; the body forces exerted on this is the
force of gravity - the weight of the liquid, what is that? Volume of the liquid - the mass
of the liquid multiplied by the acceleration due to gravity; so, what is the mass of the
liquid? rho into dy into dz; if I take unit width at right angles rho into dy dz is the
mass of the liquid - density into volume - and if I multiply it by the acceleration due to
gravity g I get the force of gravity.
So, this is the body force being exerted on this liquid; so, here is an expended view
of this, of this control volume. This is an expended view of this control volume and the
body force is rho g dy dz. What are the surface forces in the z direction? The surface forces
in the z direction are the shear stresses being exerted on the 2 vertical faces which,
the 2 vertical dz faces - these are the shear forces, these are the ones in the z direction.
So, what are the expressions for shear forces on this face? It will be mu dVz dy - this
is our Stokes law relating stress and rate of strain - and on this face, because from
this point I have moved dy, a distance dy forward in the y direction, so this mu dVz
dy becomes, gets increased by certain amount because I move a distance dy forward in the
positive y direction. So, it becomes mu into dVz dy plus d 2 Vz dy squared into dy the
whole thing multiplied by the area dz. So, the 2 shear forces acting on in the z direction
on the 2 faces - vertical faces of this control volume - are this one which is upwards and
this one which is downward. so, these are the forces, the 2 are the surface forces,
and this is the body forces acting on in the z direction on the liquid inside this control
volume dy dz.
So put all this down and we get our force balance. If I put all these force down, I
will get my force balance and I get the equation by applying; I will get, let us put that down.
What are the net z direction forces? I will get first the body force.
I will get rho g dy dz, that is the body force in the z direction, plus the shear force in
the positive direction mu d 2 Vz dy, sorry, d dVz dy plus d 2 Vz dy squared dy multiplied
by dz minus the force which is acting in the negative direction - negative z direction
- that is mu dVz dy dz. So, this is the body force, this is the second one, is the shear
force in the positive y direction on one face - one vertical face - and this is the other
shear force acting in the opposite direction on the other face of the control volume.
Newton's second law says if you neglect momentum effects the sum of all these must be equal
to zero. So, if I simplify this, the obvious, this term will cancel with this term so, if
I simplify I am simply going to get mu d 2 Vz dy squared is equal to; sorry, I will take
the mu on this side is equal to minus rho g by mu. I am simply going to get d 2 Vz dy
squared is equal to minus rho g by mu on simplification. So, the application of Newton's second law
to this control volume dy dz with the simplification that momentum effects in and out of the control
volume are negligible then gives gives me this very simple differential equation.
Let us integrate this; integrating it is a second order equation so I will integrate
it twice. I integrate it once, I will get dVz dy is equal to Vz - it is of course the
velocity in the z direction of the liquid dVz dy; it is equal to minus rho g by mu into
y plus c 1. And if I integrate it a second time, I will get Vz the velocity in the z
direction of the liquid condensate is equal to minus rho g by mu y squared by 2 plus c
1 y plus c 2, that is what I get on integration. So, I need 2 boundary conditions in order
to get the velocity profile of the liquid condensate. What are my 2 boundary conditions?
My 2 boundary conditions are at y equal to zero the no slip condition
So, at y equal to zero, Vz is equal to zero - typical fluid mechanics condition - and
at y is equal to delta. Let us now use the fact that the vapor exerts no shear stress
at the liquid vapor interface so if there is no shear stress being exerted at the liquid
vapor interface, it follows that dVz - by Stokes law - dVz dy, there is no shear stress
at the liquid interface, then dVz dy must be equal to zero. So, these are my 2 boundary
conditions - one at y is equal to zero and at one is at y is equal to delta which I put
in.
If I take both these boundary conditions, then I can show that the constant c2 - I am
now leaving out a few simple steps - you can show that c2 is equal to zero and c1 is equal
to rho g by mu delta; these are the 2 constants of the integration. Therefore, the velocity
profile of the liquid condensate is given by Vz is equal to rho g by mu multiplied by
y delta minus y squared by 2 - this is what we get for the velocity profile of the liquid
condensate inside the liquid film. This is, let me write that down, this is the velocity
distribution of, in the liquid film Vz equal to zero at y equal to zero and V is dVz dy
equal to zero at y equal to delta so this is my velocity distribution.
Let me go back to the sketch to show you the liquid, the velocity distribution here it
is on this sketch. Now, you can see the nature of the distribution that we have got is Vz
is equal to zero at y equal to zero and dVz dy z is equal to zero at y equal to delta
so we have a parabolic distribution of this form. That is the equation that we have got
for velocity distribution after applying Newton's second law of motion. Now, once I have the
velocity profile, I can calculate the mass flow rate at the condensate so what is the
mass flow rate of the condensate? Mass flow rate of condensate - if I ask you at a distance
z at a cross section z, what is the rate at which mass is flowing in the liquid at a distance
z?
Then, you will say mass flow rate using the velocity distribution, mass flow rate of condensate
at a distance z from the top - if I call that as m dot - then it will be equal to, let us
look at the sketch again now at a distance z I want to know the mass flow rate. Suppose
I ask you what is the mass flow rate flowing through this area dy and in this x direction.
It is B so B into dy is the area, the velocity is Vz and the density of the liquid is rho
so through a distance dy and an area B dy the flow rate will be rho multiplied by Vz
multiplied by B dy rho into velocity into area.
I want the value - the mass flow rate in kilograms per second - across the whole thickness delta
therefore I will integrate this expression from zero to delta to get the mass flow rate.
And if I integrate this expression I will get rho integral zero to delta. For Vz, now
I will substitute T velocity distribution that we had got. What was the velocity distribution?
rho g by mu y delta minus y squared by 2, that was the velocity distribution for Vz
into B which is a constant dy.
Perform the integration. If you perform the integration you will get rho squared g into
B - these are all constants - upon mu and I am doing the integration and putting the
limits zero to delta. We will get a 3 in the denominator and a delta cubed in the numerator.
So, the net result of the integration is rho squared g B delta cubed upon 3 mu, that is
the mass flow rate in the condensate film.
Now let us take a differential of that; if I take a differential of that, take a differential
of that, I will get d m dot by B. This is mass flow rate per unit width at right angles
to the paper is equal to rho squared g by mu into delta squared d delta if I take a
differential. So, what I am saying? Then, I take a differential; what I am saying in
effect is the following. I am saying this differential indicates that, as we proceed,
indicates this will be the result that we have got, indicates that as we proceed downwards
from a horizontal section z to a section z plus dz. If we go down a distance dz in the
positive z direction the condensate film thickness, the condensate film thickness increases from
delta to delta plus d delta with a differential amount and the mass flow rate per unit width
and the mass flow rate per unit width increases by rho squared g by mu delta squared d delta.
That is what we are saying by applying, by taking the differential of the condensate
mass flow rate or in other words what we are saying is - let me again go back to the sketch
I showed you - if I go from across the cross section z here, I know m dot by B, if I go
down a distance here the m dot by B increases by certain amount and why is that increase
occurring? Because condensation is taking place, so in other words, the condensation
rate per unit width from z to dz is rho squared g delta squared d delta upon mu or in other
words what we are saying, let me write that down in other words.
The condensation rate per unit width per unit width from z to z plus dz is rho square g
delta squared upon mu d delta rho squared g delta squared d delta upon rho squared g
delta squared upon mu into d delta that is the condensation rate per unit width from
z to z plus dz.
So, now we know the condensation, we got a velocity profile in the liquid condensate,
we have got the mass flow rate so we effectively applied conservation of mass and Newton's
second law. What remains now is to apply the energy equation the first law thermodynamics.
We will stop at this stage and next time we will apply the energy equation to get an expression
for the heat transfer coefficient and for the thickness of the condensate film.