In this lecture we are going to now see a
practical application of our knowledge of
material properties for material selection
for the heat exchanger, so this is what we
are going to see in this particular lecture.
So let us see how some of the thermal properties
that we have learned earlier can come into
picture in terms of material selection for
a very important practical system called heat
exchanger.
Now that why a heat exchanger is required
because it helps in terms of exchanging the
heat between hot and cold fluid such that
the heat flow per unit area is maximize. There
are many applications for example, in automobiles
or even in simple room heaters, in many cases
you need to actually either cool the system
very fast or you need to heat it up very fast.
In both the cases, you need to design a suitable
heat exchanger to do this particular work.
So our objective here is to maximize the heat
flow per unit area.
And what are our free variables? The tube
wall thickness and the material choice. What
are the constraints that we may actually come
across? One of them is the suppose pressure
difference, maximum pressure difference delta
p between the inlet fluid and the fluid outside.
And this is important, up to what temperature
it can withstand let us say in this particular
example, we will keep it to be sufficiently
high 150 degrees centigrade, and also corrosion
resistance can be another important point,
we will keep that point in mind when we will
finally select the material, so let us go
to the board and use the Ashby criteria to
do this same exercise.
So we have a heat exchanger, let us first
draw the exchanger. Suppose, this is the surrounding
fluid and on the surrounding fluid we have
a heat exchanger coming in and which has a
uniform diameter and suppose this part is
our boundary, so this is what is our fluid
2 boundary, this is our fluid 2. And we have
the fluid 1 inside, this is fluid 1 and we
have the temperature here is T1, the pressure
here is P1, the temperature here is T2, pressure
here is P2, so the same fluid is here all
around the fluid 2. And we also have this
one with a uniform say diameter all-around
say suppose d1.
And if we cut across this tube A here, what
we are going to see is that it has a finite
thickness, uniform thickness 
and this thickness is t of the that is the
wall thickness, right because that thickness
will become important for us when we will
analyze the system. Now, then there are three
things that we will be basically analyzing
in it, one is that there is fluid 1 inside
here right, so the first thing is the heat
that is getting convected from this fluid
1 to this solid.
Then it is conducted so this conduction and
again there is a convection, so there are
the 3 elements that will come into the picture,
so conduction, convection at the 2 other sides,
so there are these combinations of convective
and conductive heat transfer that is going
to take place in the system. Now let us assume
steady-state heat flow, let us just put our
objective of course at the top, which is maximize
heat flow per unit area.
And let us now assume the steady-state heat
flow, now the heat transfer 
between fluid 1 and the wall , so let this
be denoted as q1 which is h1 A times Delta
T1, where h1 is the heat transfer coefficient
of fluid 1 and then so we should be able to
find out, Delta T1 is of course the temperature
difference 
between the fluid and surface 1 that is the
internal surface of the fluid, the temperature
difference is going to actually is Delta T1,
so that is what is my first part A that is
the heat transfer between fluid 1 and the
wall which is happening here.
Next will be the B that is the heat conduction
and next will be C, so let us look at the
conduction. So heat transfer across the wall,
this can be written as so this is our first
equation, now the second equation is K A delta
T over t, where we know that A is the surface
area, k Delta T is known to us, t is the thickness
as I have shown here, the t is the thickness
of the wall and k is the thermal conductivity,
now the last part C, so this is my second
part.
Now last part C which is heat transfer 
between the wall and fluid 2, in that case
q3 is very similar is h 2 A Delta T 2, this
is my third equation, so we have to basically
get an equivalent expression of heat transfer
on the basis of these 3 equations that come
into the picture. So if we will keep this
particular equation in our mind then we can
go for this equivalent heat transfer now.
So let us try to erase this part now and try
to get a formulation for the equivalent one,
which takes care of all these 3 things together.
Now, one point we have to keep in our mind
here that at steady state these three, at
the steady-state 
q 1 equals q 2 equals q 3, so that has been
achieved and hence at that stage the heat
transfer the average heat transfer at the
steady-state 
is 1 over U A equals 1 over h 1 A + t over
k A + 1 over h 2 A. So it is like you have
three resistive systems and all first order
and all of them are together that is the giving
you the equivalent total resistance to the
system, so U here is the overall heat transfer
coefficient.
And it can be also written that the q in the
steady states this q equals U A delta T. Now,
what happens is that in most of the cases
this actually dominates the heat transfer,
the t over k A, and in comparison to that
we can actually neglect these two. So suppose
in any particular case it is possible for
us to do that, in that case what we can write
is that we can further simplify this expression
now and we can write it as for the domination
of heat conduction across the tube wall over
convections.
We can write that simply q equals k A Delta
T over t, right. So this we can say is the
steady-state overall heat conduction when
everything else is actually neglected. Now,
during this transfer of the fluid at pressure
P1, this wall is also subjected to some stress
which is something like a Hooke’s stress
that is happening in the wall, so it is expanding
trying to expand the wall. So from the mechanical
point of view, it has to withstand that pressure
also, so the stress that will come, from the
tube wall 
stress point of view, we can write that Sigma
has to be less than Sigma y.
And what is the stress here, that is Delta
P, the pressure difference times r, the radius
of the tube divided by t that has to be less
than or equal to Sigma y. In fact, by keeping
thickness as the free variable here, so thickness
is here as a free variable, so I can write
that t equals Delta p times r let us call
it as r1, r1 equals our d1 by 2, so Delta
P times r 1 divided by Sigma y. Now I can
substitute this in this equation of overall
heat transfer so that means I can write here
that q equals k A Delta T times Sigma y divided
by Delta P times r 1, right.
So I can write also as q equals to, let us
keep all the other parameters separate that
is A Delta T over Delta P r 1 and k Sigma
y. So in a heat exchanger, if I want to increase
the heat exchange rate. more heat to transfer
from fluid 1 to fluid 2 so that either I can
heat up fast or I can cool fast, I must increase
this k times Sigma y that is both k and the
yield strength has to be increased, so that
it can sustain the pressure, at the same time
the higher thermal conductivity allows us
to take more temperature, to take more heat
from fluid 1 to fluid 2. Now let us look into
that what are the possibilities that are there
for us for this particular system.
So now if you look into the Ashby chart, we
will see that in this particular case, so
if you take a logarithm of the two that means
if k Sigma y is a constant, then log of k
would become - log of Sigma y + log of C,
so that means there is a negative slope and
that negative slope considering the 150 degree
temperature difference will bring us in this
particular line.
This is a line with negative slope and if
this particular line now we actually look
at it, so this is a curve this is relationship
between thermal conductivity and . With this
negative relationship all I have to see is
actually see everything beyond this point
because everything below this point are not
suitable for the heat exchanger design here
up to 150 degree centigrade, so if I magnify
this part what are the possibilities for us
that is coming?
You can have the brass UNS alloys; we can
have aluminium bronze alloys, phosphor bronze
alloys, nickel base alloys, silicon bronze
alloys, so these are the possibilities that
come in our particular slide area. So if I
look at it further, you will see that the
brass is like naval brass here the k Sigma
y index is the highest 5 into 10 to the power
4, the only problem is for naval application
that is a dezincification that means it is
susceptible to the environmental pollution,
so that is why the corrosion point of view
this is not a good idea.
Phosphor bronze, 4 into 10 to the power 4
which is cheap, but not as corrosion resistant
as the next one that is the aluminium bronze,
which is slightly lower 3 point 8 into 10
to the power 4, but it is economical, it is
corrosion resistant and this comes out to
be a more practical choice than the brass
or the phosphor bronze from the corrosion
point of view. Nickel, iron, aluminium bronze
is actually more corrosion resistant, but
it is more expensive and this also is little
less. Similarly, silicon bronze is less good
because its index is less.
So out of all the choices as we can see that
the first three are better choices from the
material index point of view, but if we put
corrosion as the constraint and also economy
as the constraint, then aluminium bronze wrought
is coming out to be the best choice and that
is what is generally used in the heat exchanger
design. In the next lecture, we will learn
about the electrical properties of the system,
thank you.
Keyword- heat exchanger, Ashby approach
