Tip:
Highlight text to annotate it
X
Last time we have started with the topic of boiling and I, first of all, I said I described
to you when boiling will occur, then I classified two types, two types of boiling; I said there
is pool boiling and flow boiling and then we focused our attention on pool boiling and
I identified various regimes or various regions in pool boiling and I drove a typical pool
boiling curve while identifying those regions. What were those regions? They were first of
all the natural convection region, then the nuclear boiling region and the third one was
the film boiling region. Even in the film boiling region, there were 2 parts, one part
which I called the transition boiling region in which we have a transition from nucleate
to stable film boiling and the second part which are called as stable film boiling. I
also identified one important point on that pool boiling curve and I said that point is
called that peak heat flux or the critical heat flux and I told you why that point is
important.
If heat flux is the parameter that is under your control and if it is the quantity that
is varying then if it goes past the critical heat flux, then the temperature difference
Tw minus Ts - wall temperature minus the saturation temperature - will shoot up from a value which
might be 10 or 20 degree centigrade to perhaps a few00 degree centigrade, even may be a thousand
degree centigrade, and there is a possibility that the heated surface may get damaged. And
that is why whenever heat flux is the controlling factor and it is in most in many situations,
then one would see that the peak heat flux is that exceeded in order to prevent any damage
to the heating surface.
Now, today we are going to look at a number of correlations equations to use in doing
calculations in pool boiling. We will focus our attention on saturated pool boiling correlations
in saturated pool boiling.
So, today we are going to look at correlations in saturated pool boiling. By saturated pool
boiling I mean that we have of course a pool of liquid and in that pool of liquid the water
or the liquid - whatever it is - is at a certain pressure and at a temperature, saturation
temperature corresponding to that particular pressure. So, that is what we call as a saturated
pool. It is at a particular pressure and it is also at the particular saturation temperature
corresponding to that particular pressure. We add heat to that liquid just as I had indicated
in the sketch last time. We go on adding heat and we will go through the various regions
of boiling and in those various regions, what are the correlations to use?
So, the first region which we have identified is natural convection that is region one.
Now, in region 1, I have mentioned it but let me repeat, change of phase does not occur
at the heated surface. That is a point which I made to you last time. The heated surface
is at the bottom; the change of phase doesn't occur there. What happens is, because of the
small temperature difference the water gets slightly superheated water or the liquid - whichever
we are talking about - get slightly superheated may be by a degree 2, degrees or something
like that. The superheated liquid will obviously have a lower density; it is going to rise
and will set up a natural convection pattern in that pool, in that pool of liquid. The
superheated liquid then eventually will reach the free surface of the pool and at that free
surface you will get evaporation. So, that is the nature of this region.
Obviously, if there is no change of phase, no vapor formation inside the liquid, it is
clear that all you have is a natural convection single phase phenomenon taking place with
superheated liquid and therefore it is also obvious that all the correlations which we
have studied earlier in single phase natural convection will be valid in this region. So,
that is the point I want to make first of all that all single phase - let me write that
down - all single phase natural convection correlations are valid in this region.
So, there is nothing new to do. You can go back to our chapter on single phase natural
convection, look up the various correlations we have got. Correlations, for example, for
horizontal tubes - they have a correlation for a heated surface facing upwards; we have
correlations for other situations are also available in the literature. Anyone of them
is valid depending upon the heated surface that we are talking about in this particular
situation.
For example, let us say we have got a situation; as an example, let us consider that we have
horizontal wires or cylinders which are being heated in a pool - horizontal wire, horizontal
cylinder which is inside a pool of liquid. Then, if boiling is going to take place and
it is the natural convection region, the relationship between heat flux and temperature difference
Tw minus Ts are as follows. Now, I am going to read these out because you have done these
correlations earlier in natural convection. What is the difference between what the way
these are written and what was written earlier in natural convection? It is when we were
studying natural convection, these relations were given in terms of a Nusselt number as
a function of Rayleigh number and Prandtl number.
Now following the practice adopted in boiling, we have written these relations in terms of
heat flux equal to something multiplied by temperature difference multiplied by the particular
relationship. So let me just illustrate what I mean; what I mean is the following. Earlier,
the relationships were of the form.
NuD is some function of Rayleigh number and Prandtl number; these were how the relationship
was given earlier. Now what is Nusselt number? Nusselt number is nothing but hD by k which
is a function of Rayleigh number and Prandtl number and h is nothing but q by A divided
by Tw minus Ts multiplied by D by k equal to a function of Rayleigh number Prandtl number.
So, these relationships which are, which I am showing you are written in the form q by
A is equal to k by D multiplied by Tw minus Ts and multiplied by the functional relationship
which is available from the correlation in natural convection. This is how these relationships
which I am showing you are put down. It is, they are equivalent obviously as I had shown
to the relationship given earlier.
So, let me again show these now. If the geometry is a horizontal wire or cylinder, the relationship
between heat flux and temperature difference are as follows. Properties again as previously
would be evaluated at the mean film temperature Tw plus Ts by 2. If the geometry something
else, say for example heated plate facing upwards. Go back to the single phase natural
convection chapter, look up the correlation there. Nusselt number is a function of Rayleigh
number and Prandtl number, whatever it is. And convert it into this form q by A equal
to k by L into Tw minus Ts into something. So, you should be able to write down such
relationships depending upon the particular geometry you are studying. I have given one
as an example; now let us move on to region two. In region 2, we have nucleate boiling.
Nucleate boiling means bubbles now form at the heated surface.
It is now a situation in which, a typical situation is - at particular sites on the
heated surface, a bubble is formed, it grows in size, then when it grows to a certain particular
size because of buoyancy forces, it detaches from the surface and rises. Another bubble
forms at the same size, again grows in size and again rises, and this phenomenon takes
place at a number of sites. As you go on increasing the temperature difference the number of sites
at which this happens goes on increasing. The frequency with which the bubbles form
per minute at a particular side also goes on increasing, so that is the nature of nucleate
boiling and we have described it last time.
Now many correlations have been developed by various investigators over the years for
this particular phenomenon and I am going to give in this lecture one particular correlation.
The correlation that I am going to give is a correlation given by Rohsenow; it is called
the Rohsenow correlation quite appropriately. Now, let me just read out the correlation.
Then, I will describe what are the parameters in this correlation. The correlation is on
the left hand side - Cpl multiplied by Tw minus Ts upon lambda that is on the left hand
side, this is a dimensionless parameter.
And you know the symbols, what they mean? Cpl is going to be the specific heat of the
liquid, Tw minus Ts is the temperature difference between the wall and the saturation temperature
of the liquid divided by lambda which is the latent heat of vaporization. It is a dimensionless
number making up the left hand side of the equation equal to Ksf - Ksf is a constant
which I will talk about in a moment - again multiplied within square bracket by a quantity
q by A, that is the heat flux being supplied, divided by mu l lambda. mu l is the specific,
is the viscosity of the liquid; lambda is the latent heat of vaporization. Within the
square root sign square root of sigma upon g rho l minus rho v; sigma is the surface
tension and g is the of course the acceleration due to gravity; rho l - the density of the
liquid, rho v - the density of the vapor. The quantity within the square bracket is
also a dimensionless number, keep that in mind. It is often called also some kind of
a Reynolds number - bubble Reynolds number. It is sometimes referred to as a bubble Reynolds
number; one can attach some physical significance to it. And then the third dimensionless number
is the Prandtl number Pr; Prl which is the Prandtl number of the liquid.
So what Rohsenow did was to correlate data for a wide variety of liquids which had been
obtained by various investigators and show that these 3 dimensionless parameters which
I have mentioned could be correlated through an equation of this form. Now let us talk
a little about this correlation; first of all, again let me go back to the symbols.
I will not go through these in detail because I have taken them up one by one.
Cpl - the specific heat of the liquid, Tw minus Ts - the temperature difference between
the surface and the liquid saturation temperature, lambda - the latent heat of vaporization,
q by A - the heat flux in watts per meter squared, mu l - the viscosity of the liquid,
sigma - the liquid vapor surface tension, rho l - the density of the liquid, rho v - the
density of the vapor; these are known symbols to you. Prl - the Prandtl number of the liquid,
n the constant at the end, Prl to the power of n in that equation is an exponent on the
Prandtl number. From the experimental data, it is found that for water the data fits best
if n is taken as one. For other liquids, the experimental data fits best if n is taken
as 1.7. So, n has 2 values depending upon the liquid which is being boiled and finally
we have a constant Ksf, a constant whose value, it is not a whole value, whose value depends
upon the surface liquid combination. And the value of Ksf was determined by Rohsenow and
by others for a variety of surface liquid combinations. So, let me just show you a table
in which values of Ksf are tabulated. Here is a table; it is available in various books.
I have just given and there are more values of Ksf for other surface liquid combinations
as well. So, on the left hand column is surface liquid combination, the right hand side is
the value of Ksf as obtained by correlating the experimental data. What do we have? Typical
values if the surfaces is copper and water is the liquid being boiled .013, for platinum-water
.013, for nickel-water .006, for brass and water .006. Then, there are 4 values here
for stainless steel and water with the stainless steel surface being subjected to different
surface treatments.
So, the first one is stainless steel and water with the stainless steel surface mechanically
polished .032, stainless steel and water where it is chemically etched .0133, third one stainless
steel and water where the stainless steel is coated with Teflon - .0058 and finally
stainless steel and water where it is ground and polished .0080. Notice that as the stainless
steel surface becomes smoother which is the case in the last 2 surfaces. Compared to the
first 2 there, the first 2 it is mechanically polished or chemically etched, the last 2
it is Teflon coated and ground and polished. Notice that in the last 2 cases the value
of Ksf is about half compared to the value earlier. So, the value of q by A for, thus
the value of Tw minus Ts for the same q by A will be half the value that you get for
the mechanically polished surface. Not half, I shouldn't say that because there is the
cube one third power on the q by A but it will be less. So, these are typical values
of Ksf and there are values of Ksf for other surface liquid combinations as well.
Now in this correlation, let me make some remarks on this correlation. First of all
the correlation was developed for horizontal wires.
The Rohsenow correlation is used for horizontal wires, horizontal tubes, horizontal plates. All properties are evaluated
at the liquid saturation temperature
and the value of constant Ksf - I have already discussed what it means and shown you typical
values. Keep in mind that the Rohsenow correlation, what it is basically saying is that q by A,
if I cube both sides of the Rohsenow correlation, what the Rohsenow correlation is saying is
for a given liquid, for a given surface and a liquid, the q by A is proportional to Tw
minus Ts to the cube. That is what the Rohsenow correlation is saying; it is useful to keep
this dependence in mind.
Now, let us go on to the next point. On the curve the nucleate boiling region ends when
we reach the point A on the pool boiling curve which I mentioned to you last time and what
is that point A? The point A is what we call as the critical or the peak heat flux. Now
for this particular point a number of equations have been developed to calculate the critical
or the peak heat flux. I will give you a few of them.
First one that I am going to give you is a correlation; an equation I should say developed
for an infinite horizontal plate facing upwards that is something like this. I have a plate;
let me draw it on separate paper.
The first equation is for a plate like this which is heated facing upwards and receiving
heat and there is the pool of liquid above it like this and it is supposed to be infinite
in extent. Based on a stability analysis, a hydrodynamic stability analysis, Kutateladze
and Zuber derived an expression for the peak heat flux for an infinite horizontal plate
which is of this form - q by a critical is equal to pi upon 24 lambda rho v multiplied
by sigma g rho l minus rho v upon rho v squared to the power of 1 by 4, multiplied by rho
l plus rho v upon rho l to the power of half. It is a classical expression derived based
on hydrodynamic stability considerations. All properties are evaluated at the saturation
temperature Ts of the pool.
Now, for most situations the second term rho l plus rho v divided by rho l to the power
of half, for most situations this second term here tends to be almost unity. And if you
put that simplification in, then the expression reduces to the simpler form q by A is equal
to pi upon 24 lambda rho v in the square bracket sigma g rho l minus rho v upon rho v squared
the whole thing to the power of one fourth. This would be the value of the critical heat
flux and this expression is very widely used. There is a slight modification to this expression
which was derived few years later and I am going to give you that. If the plate is finite
sized, not very small but finite sized, then
For a finite sized horizontal plate, Lienhard and Dhir showed that the expression was slightly
modified. Instead of having a pi by 24, they derived a constant .149; the rest of the expression
is the same that we had earlier. So, q by A derived by them was: q by A critical is
equal to .149 lambda rho v to the power of half multiplied by sigma g rho l minus rho
v the whole thing to the one fourth power. So, we use this expression or the earlier
one depending upon which one we prefer. There is a 10 percent difference or so between the
two. This expression derived by Lienhard and Dhir is valid so long as the plate is not
too small or shall we say large and the way to show it is to calculate the dimensionless
quantity L into bracket g rho l minus rho v upon sigma to the power of half and to show
that this quantity is greater than 2.7 where L is the characteristic dimension of the plate.
So, what one finds for typical liquids is that unless the plate is extremely small,
a few millimeters or so, this expression generally you will get a value greater than 2.7. So,
for calculating peak heat flux the preferred expression is the expression given by Lienhard
and Dhir which is q by A equal to .149 lambda rho v to the half into bracket sigma g rho
l minus rho v the whole thing to the power of one fourth. This is the expression that
is now a days preferred; there are expressions for other geometries also. I am not giving
them here but there are expressions available for other geometries like horizontal tubes,
etcetera for the peak heat flux.
Now, finally we come to the boiling region; I mean the film boiling region. First, we
have the transition boiling.
And you recall my mentioning to you the transition are, sure describing to you that transition
boiling is the region in which Tw by the q by A value decreases as Tw minus Ts increases.
And this particular thing happens, this situation occurs because it is a region in which part
of the surfaces covered by a film and part of the surface nucleate boiling is occurring.
And when the film boiling takes over suddenly then the vapor having a low thermal conductivity
puts an additional thermal resistance in the path of the heat flow. So, for a certain Tw
minus Ts you require, you get a lesser q by A flowing across the vapor film. That is why
one gets a decrease in the q by A with increasing Tw minus Ts.
Transition boiling is not something that is widely studied and the reason is it is not
something that occurs that often in practice. It has been studied by few investigators.
There are some paper showing curves of what you get for different liquids and surfaces
but no satisfactory correlation which can go across various sets of data has ever been
obtained. So, we will not give any correlation at this in this particular for this particular
phenomenon.
The next in the same region is the stable film boiling region and here for stable film
boiling we will give a correlation which was derived first by, give an expression which
was derived first by Bromley. The particular expression which I am showing you is valid
for a horizontal tube so Bromley derived this expression for the geometry of a horizontal
tube something like this. That means we have a horizontal tube
Let me draw one draw across section of the horizontal tube. Let us say this is the horizontal
tube sitting in a pool of liquid, this is the pool of liquid. Sitting in a pool of liquid,
heat is being supplied like this to the pool of liquid and the temperature difference is
large enough that we have got stable film boiling. So, what will happen? What will happen
is you will get a film of vapor all around that tube forming something like this. Let
me draw a film. I will draw an exaggerated thickness again; we will get a film of vapor
like this forming around the tube flowing upwards because the density of the vapor will
be less than that of the liquid. So, this is vapor flowing upwards and then forming
some kind of a pattern as it goes out of the liquid upwards. So, this is what we will get
if you have got stable film boiling - a film of vapor all around the heated surface, the
film of vapor rising and then going up as a flume at the top. That is what you will
get so this is what we are talking about.
Now, notice the similarity between this and what we saw in film condensation. There we
had a film of liquid flowing downwards now we have a film of vapor flowing upwards. One
has to flow downwards because the liquid was heavier; here this has to flow upwards because
the vapor is lighter but keeping this similarity in mind Bromley derived an expression for
the heat transfer coefficient. Basically, Bromley did a derivation very similar to what
the Nusselt derivation was like which we talked about when we studied condensation. So, you
see the similarity of the expression now. The expression obtained by Bromley was hc
is equal to .62 into bracket lambda prime lambda dash into rho v into rho l minus rho
v gkv cubed divided by Tw minus Ts mu v into D, the diameter of the tube, the whole thing
to the power of one fourth.
You will see it is very similar to the expression we had in condensation which was lambda rho
squared gk cubed upon Tw minus Ts Ts minus Tw into mu D. Very similar excepting that
instead of the properties of the liquid, now you have the properties of the vapor. You
have mu v kv rho v and then you have rho l minus rho v because you have buoyancy. You
have the liquid moving, the vapor film moving upwards. You also have, instead of lambda
you have lambda dash and what is lambda dash? lambda dash is the difference between the
enthalpy of the vapor at the mean film temperature Tw plus Ts by 2 and the enthalpy of the liquid
at saturation temperature. This is the definition of lambda so it is not just latent heat of
vaporization it is a latent heat plus an enthalpy of vapor associated with superheat of the
vapor. That has to be taken into account.
Now, when you have a film of vapor surrounding that tube and the temperature difference Tw
minus Ts now is going to be high; if you recall the curve I showed you last time I mentioned
that in stable film boiling, the value of Tw minus Ts can be a few00 if not a few thousand.
Let me go back and show the step, typical pool boiling curve of last time.
We have, talking of this region, the stable film boiling region where Tw minus Ts is in00s.
When you are in this region and you have a vapor film, obviously radiation across the
vapor film is also going to be an important factor and we have to therefore take that
into consideration. So, what Bromley did was doing a Nusselt type of analysis, he derived
what he called as a convective heat transfer coefficient.
Then he went onto say radiation is also important and said since radiation is significant in
film boiling because of the high values of Tw minus Ts, the radiation component has to
be added on in order to obtain the total heat transfer. And he showed that the total heat
transfer coefficient is the convective heat transfer coefficient hc plus 3 by 4 times
the radiative heat transfer coefficient. Whenever hr is less then hc, he showed that his expression
is valid within the accuracy that one wants.
We will not go into the details; I am just asking you to accept this. I am saying calculate
the convective heat transfer coefficient from the expression which Bromley gave earlier
which I have shown you that is this expression. Then, calculate the radiative heat transfer
coefficient by using formulae for radiant heat exchange between parallel planes taking
the emissivity of the liquid to be unity. So, hr is the radiation heat transfer coefficient
calculated from the formula for radiative heat exchange between parallel planes taking
the emissivity of the liquid to be unity and then say that the total heat transfer coefficient
is hc plus 3 quarters hr. That is what I am asking you to do; so this is how one would
calculate the heat transfer coefficient in stable film boiling. All vapor properties
are evaluated at the mean film temperature; all vapor properties means mu, v, kv, etcetera
that we need.
All vapor properties are evaluated at the mean film temperature; the liquid density
is evaluated at the saturation temperature. So, this set of equations, the convective
heat transfer coefficient and the radiation heat transfer coefficient appropriately added
on gives us the total heat transfer coefficient. Now, let us do a few problems to illustrate
the use of these correlations which I have given you. We will first of all do the following
problem.
Problem is the following, we say - there is a pool of liquid A; let me read it out - A
.14 centimeter diameter horizontal platinum wire is held in a pool of saturated water
at 100 degree centigrade and 1.013 bar. This 100 degree centigrade corresponds to this
saturation pressure. The wire is electrically heated so that boiling occurs. Calculate q
by A when the temperature difference Tw minus Ts is 1 degree centigrade. Calculate Tw minus
Ts when the temperature difference is 15 degrees centigrade; that is the problem.
So, let me just draw a pool, let us say we have a pool of liquid like this. This is our
pool of liquid like this and we are told that it is at 100 degree centigrade, Ts. It is
a saturated pool at 100 degree centigrade. In this pool, I have a horizontal heated platinum
wire. Let us say this is the wire shown in red here. This is the platinum wire
which is electrically heated; I would like to know for Tw minus Ts and Tw is the temperature
surface temperature of this wire, platinum wire at Tw. So, the problem is if Tw minus
Ts is 1 and 15, what is q by A for these 2 values of Tw minus Ts? That is the problem.
It is a straight forward substitution into the correlation for natural convection and
for nucleate boiling. So, let us do the problem now.
Let us, in the first case we will make the assumption for Tw minus Ts - let us do that
first - equal to 1 degree centigrade. We will assume that we are in the natural convection
region which is a good assumption; up to a few degrees this is usually true. We are in
the natural convection region in pool boiling so if I am to use a natural convection correlation,
I will use the single phase correlation for horizontal wires which I showed you earlier.
What do I need for doing calculations now? I need first the mean film temperature; now
systematically, I will go about the calculation. I will say mean film temperature is equal
to 100 plus 101 by 2 which is equal to 100.5 degree centigrade. The properties of water
will be, which if you look up you will get at the mean film temperature, would be beta.
You need the expansion coefficient, you need the Kinematic viscosity, you need the thermal
conductivity and you need the Prandtl number for substituting into the correlation. The
values - if you look up in a table, you will get the following. Beta Kelvin to the minus
1 nu is .294 into 10 to the minus 6 meter squared per second, thermal conductivity - .683
watts per meter Kelvin and the Prandtl number of the liquid - 1.74. These are properties
which you look up for water at 100.5 degree centigrade. Calculate the Rayleigh number
now.
So, you get the Rayleigh number RaD is equal to GrD Pr which is equal to g beta Tw minus
Ts D cubed divided by nu squared - this is the Grashof number with D as the characteristic
dimension divided by nu squared the whole thing. say Grashof number multiplied by Pr.
Substitute the numbers which we have got and I am skipping the substitution step; you will
get in this particular case 409.1. Now, go to the correlation for natural convection
and you will get q by A is equal to .683, which is k for water, divided by D .14 into
10 to the minus 2. That is the diameter of the wire into Tw minus Ts 101 minus 100 multiplied
by the correlation is .36 plus .518 multiplied by the Rayleigh number to the power of one
fourth the whole thing divided by 1 plus .559 divided by the Prandtl number 1.74 to the
power of 9 by 16 the whole thing to the power of 4 upon 9.
This is the correlation for natural convection and just substituting it is nothing very deep
and if you calculate this, you will get 1117 watts per meter squared so this side the first
answer that we are looking for. We are saying for a temperature differential of Tw minus
Ts equal to 1, we are likely to be in the natural convection region. So, we have used
natural convection correlation for single phase for a horizontal wire and got the q
by A which we would need. In the heat flux we were needed to supply for this particular
temperature difference.
Now let us go on to the next part; the next part is calculating q by A for Tw minus Ts
equal to, for Tw minus Ts equal to 15 degrees, that is the second part. Quite obviously,
we are now going to be in the nucleate boiling region; for this particular Tw minus Ts assume
nucleate boiling. Nucleate boiling usually starts 3, 4, 5 degrees centigrade for water,
may go up to 25, 30 degrees for water. Assume nucleate boiling; now we will need, for using
the rho, we will need the Rohsenow correlation and in order to use the Rohsenow correlation
we will need properties at the saturation temperature 100 degree centigrade. And what
are the properties? I will just put down the values and then just say - what the answer
will be. You will get, if you look up the properties of water at 100 degree centigrade,
you will get Cpl equal to 4220 joules per Kelvin, joules per kilogram Kelvin, mu l the
viscosity equal to 282.4 into 10 to the minus 6 kilograms per meter second, sigma the surface
the liquid vapor surface tension 589 into 10 to the minus 4 Newtons per meter, density
of the liquid equal to 958.4 kilograms per meter cubed, then we will need the density
of the vapor rho v.
So, we will get from the table rho v density of the vapor .598. If this is not available
in the liquid tables, of giving the properties of the liquid, you may have to go to steam
tables. .598 kilograms per meter cubed and the Prandtl number of the liquid equal to
1.75. You will also need the constant Ksf so go to the table which I showed you. The
constant Ksf is, for this particular case platinum and water, .013 from the table. So,
we have got all the values. Now, it is a straight forward substitution into the Rohsenow correlation;
substitute into the Rohsenow correlation.
I am now skipping the steps; I am not going to substitute all these numbers but if you
do that you will get q by A is equal to 141.32 multiplied by 115 minus 100, this is the temperature
difference, Tw minus Ts the whole cubed and that comes out to be 476955 watts per meter
square. Notice the steep increase in the value of q by A when Tw minus Ts goes up from 1
to about 50. We have gone up by an enormous factor; earlier the q by A in natural convection
was only about 1000 now it is 476000. And just to illustrate that, let us go back again
look at the pool boiling curve.
Notice it is a log plot here; this is the natural convection region so our first point
was somewhere here, which we are calculating some region here. Now, we are calculating
a point which is somewhere here and it is a steep rise where q by A is proportional
to Tw minus Ts to the q power. So, our first point that you have calculated is somewhere
here in this problem; the second point that we are calculating is somewhere here in this
problem. Now, so this is our second answer; these were the 2 answers we were looking for.
Use appropriate correlations to calculate q by A for different temperature differences.
Now let us do one more problem in film boiling so that we illustrate the use of Bramley's
equation.
The problem we are going to do is the following, the problem is - an aluminum cylinder, 2 centimeters
in diameter and 15 centimeters long, is heated to a temperature of 500 degrees centigrade
and immersed horizontally in a liquid nitrogen bath at minus 196 degrees centigrade. Liquid
nitrogen has a saturation temperature of minus 196 degree centigrade at around atmospheric
pressure. Neglecting heat transfer from the end faces, calculate the initial heat transfer
rate. Take emissivity of the aluminium surface to be .4. This is the problem which we want
to do next; now here is the solution.
The first thing to notice - the surface of the aluminum cylinder which is heated is at
500. The liquid nitrogen is at minus 196 so when you immerse the cylinder in the liquid
nitrogen horizontally you have got a temperature difference initially of 696 degrees centigrade.
At that temperature difference, you are going to get stable film boiling, that is what you
are going to get. So, that is the first thing to recognize; so we are going to use the equations
for stable film boiling.
So, here is our pool of liquid which I am showing in this sketch; here is the heated
aluminium cylinder at 500 and this is the saturated liquid at minus 196 so stable film
boiling is going to occur. So, if I look in the cross section of this, this is my aluminium
cylinder and around it a film of vapor is going to form like this and rise to the surface.
A flume is going to rise to the surface. We want to calculate the heat transfer coefficient;
it is again a straight forward substitution so let us go through the whole procedure.
The first thing to know - it is the mean film temperature at which we will need properties.
Mean film temperature of the vapor equal to 500 plus minus 196 by 2 which is 152 degree
centigrade. So, we will need properties of the vapor at this temperature. This case,
the particular vapor is nitrogen. It is liquid nitrogen so nitrogen gas, we have to look
up these properties, has the following properties. You have to look them up in the property tables
for nitrogen; it has the following properties at 152 degree centigrade. You have kv the
thermal conductivity .0349 watts per meter Kelvin, rho v is equal to .80 kilograms per
meter cubed, mu v the viscosity 23 into 10 to the minus 6 kilograms per meter second.
Also we will need the Cpv; the average value of this specific heat of the vapor which is
.10 not point sorry 1048 joules per kilogram Kelvin. We will also need the density of the
liquid at minus 196 which is 800 kilograms per meter cube and we will need the latent
heat of vaporization for nitrogen at that particular temperature which is 201 into 10
to the power of 3 joules per kilogram. One has to look up all these properties in appropriate
property tables; I am giving you the values here. Now, we calculate first of all lambda
prime; lambda prime is the latent heat plus the enthalpy associated with the superheat.
So, 201 into 10 to the power of 3 plus the average specific heat of the vapor multiplied
by 152 minus 196, that is at the mean film temperature, and this will come out to be
565.7 into 10 to the power of 3 joules per kilogram. Then, substitute into the Bromley
equation and now I am not going to do the substitution; I am just going to give you
the result. If you substitute into the Bromley equation, substituting into the Bromley equation
I am skipping putting down the details of the numbers, you will get hc is equal to 91.34
watts per meter squared Kelvin. So, I want you to do this calculation on your own with
the property values I have given.
Now, you have to calculate hr; calculating hr means I need to know the heat, the flow
of heat, by radiation - q by A radiation. I use the formula for parallel plates, q by
A by radiation is the Stefan Boltzmann constant 5.67 into 10 to the power of minus 8, Tw to
the power of, 4773 to the power of 4 minus 77 to the power of 4 divided by 1 upon .4;
.4 is the emissivity of the aluminium surface, 1 upon 1 one is the emissivity of the liquid
minus one, which comes to be 8096.9 watts per meter square and now from this I calculate
hr so I will get hr is equal to - the heat flux in radiation point 8096.9 divided by
the temperature difference 773 minus 77 which comes out to be 11.63 watts per meter squared
Kelvin.
Therefore, the total heat transfer coefficient is hc plus 3 quarters hr which is 91.34 plus
3 quarters into 11.63 which is equal to 100.07 watts per meter squared Kelvin. This is the
total heat transfer coefficient and therefore the initial heat flux or the initial heat
transfer rate rather - that is what we are to calculate. Initial heat transfer rate is
equal to the heat transfer coefficient multiplied by the temperature difference 773 minus 77,
this gives the heat flux, multiplied by the area pi into .02 multiplied by .15 and this
comes to 656.4 watts.
We are neglecting as we have told neglect the heat transfer from the end faces from
the horizontal cylinder. So, this is the answer we are looking for; we have done it in steps
remember. We calculated first the convective heat transfer coefficient using the Bromley
equation, then we calculated the radiation heat transfer coefficient. We added them up
with the formula given and then got the total heat transfer rate. Now, let us just sum up
for one minute what we have done in this topic in condensation and boiling.
Under condensation, we did drop and film condensation. We did film condensation on a vertical plate
and film condensation on a single horizontal tube and a bank of horizontal tubes. In boiling,
we first described what is pool boiling and flow boiling. We discussed a typical pool
boiling curve; we then did various types of pool boiling while we were discussing the
curve. And then gave correlations in the saturated pool boiling region for natural convection
region, for the nucleate boiling region, for the stable film boiling region and correlation
for the peak heat flux and finally we did some problems today to illustrate the use
of these correlations. With this we have come to the end of the topic of condensation and
boiling.