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Welcome to lecture four of module two on heat transfer. We are continuing our discussion
in the area of one-dimensional and steady state heat transfer in this lecture. We will
start discussing on extended surface heat transfer. From the name itself you can understand,
that heat transfer surface is extended, that means, we have increased the surface area.
So, the heat exchange phenomenon, that is, what we are going to study in this lecture
by increasing, in increasing the surface area of the heat exchanger.
Now, the question is that why this extended surface is necessary? We understand, that
heat transferred is accompanied by cooling a gas or heating a gas or cooling a liquid
or heating a liquid and so, heat transfer phenomena is accompanied with the solid surface
and the, and fluid medium where heat transfer coefficient is involved, and it has been found,
that mean, certain situation the heat transfer coefficient is very poor. And then, for practical
applications, practical necessity we may have to increase the rate of heat transfer and
therefore, extended surface heat transfer concept has come.
Now, if we just think about that, in case of any convective heat transfer we have told
that rate of heat transfer is equal to convective heat transfer coefficient into area of heat
transfer into the driving force. Now, so what happens is that if you have to increase the
heat transfer rate, then what we have to do is, we have to increase either the heat transfer
coefficient or heat transfer area or the driving force.
Now, when we can tell about that deriving force, the delta t, delta t is normally or
usually, it is fixed by the extended wall. For example, you have to heat a fluid up to
a certain temperature, say suppose, pre-heating a liquid stream for additional column. So,
that is already being decided how much heat transfer is needed for this kind of situation.
Similarly, you have some hot fluid coming from a say, flue, flue gas chimney and you
want to take, extract, that heat. So, you know the temperature of the, of the flue gas.
So, the differential temperature also, is mostly fixed by reaction wall or by the problem
of concern. Therefore, we have little scope in doing something on delta t part to increase
the rate of heat transfer. Now, the, the other part is, say, heat transfer
coefficient. Yes, we know, that heat transfer coefficient or convective heat transfer coefficient,
this value is very small, particularly for gases and sometimes it is also not very high,
even some liquids also, but actually, this values of H is governed by the hydrodynamic
conditions of the fluid, which are getting heated or which are getting cold.
So, therefore, it may so happen that we have to increase the value of H. If we try to increase
the value of H, how much we can? We cannot increase infinitely the value of H because
it is governed by the hydrodynamic conditions, say for a, for example, if you have a liquid
at laminar flow. If you just convert in turbulent flow situation, you can increase the heat
transfer coefficient by, to some extent. We will see in future when we will be discussing
in details about the convective heat transfer, that heat transfer coefficient by, can be
increased by changing the velocity, by increasing the velocity or by changing some hydrodynamic
conditions. But the increase have, the increase in the heat transfer coefficient has some
limited, is limited, it is not indefinitely it can be increased.
Then, it also happens, that for gas-gas system it is hard to increase to the heat transfer
coefficient for this kind of, for gases heat transfer between gas and gas with the help
of a solid barrier. So, from the gas to the solid and solid to the gas, this is very hard
to increase the heat transfer coefficient for this kind of situations.
So, what is needed now? Then, another option is left with us, which is nothing, but the
heat transfer area and this heat transfer area, yes it is mostly dependent on the designer
who is designing the heat direction equipment. Therefore, the concept of extended heat transfer
has come into picture. So, one can increase the heat transfer area by putting some kind
of extended surface on to the basic equipment and that way, that heat transfer rate can
be substantially increased, it can be several hundred times increase is possible. We will
see in, in successive or in, in, in, in, in, subsequently we will see, that how the heat
transfer coefficient, that increase.
Now, if we have to have, as I discussed, a, a quicker heat transfer, in many situations
we may be requiring, then we need to have the, the most important option at hand is
that increasing the surface area for heat transfer. And this increasing the surface
area of that is that nothing, but the extended surface, this is obtained by attaching a rectangular
metal strip or annual rings, which is called as fins, to the surface of heat transfer.
And the typical application areas for these fins are heat transfer for transformers and
motors, IC engines of motor cycle and various heat exchangers. They use this extended surface,
heated, heat, heat transfer phenomena. And the various types of fins, which are used
for different situations are the individual rods known as pin fins or spines and they
can be cylindrical spines, truncated conical spine, parabolic spine. Then, continuous blades
known as fins and then, these fins can be longitudinal fins or radial or circumferential
fins with various profiles, such as rectangular, trapezoidal, parabolic, truncated, conical,
etcetera. So, what we would see is in the next slide
we will try to have look into different types of fins, that is available and that is being
used.
So, you see that there are some pictures we have discussed. There is that longitudinal
fin of rectangular profile, then longitudinal fin of trapezoidal profile, you see, this
is the trapezoidal profile here; similarly, longitudinal fin of parabolic profile. We
can understand, that this, there is curvature and this is nothing, but the parabolic profile.
Then, we have a circular tube with fins of rectangular profile. Again, this is a rectangular
profile. We can see, that this, this is, particularly this is a rectangular profile, then this is
the rectangular profile, then here is cylindrical tube equal to the radial, also called circumferential
fins. And you can see, that it is the, this is cut here to see the profile; here, it is
a rectangular profile again. Similarly, it can be a truncated conical profile like this.
Then, we can have several other kinds of, like spines, which is just simply a rod protruding
outward, that cylindrical spine it is thorough, thoroughly, that its, its area is same. Truncated
conical spine, so its area is continuously changing and also, we have parabolic spine
here, the surface area is also changing, not only that, the curvature is also, there is
a parabolic curvature. So, we can understand certain things that
fins of, could be of a longitudinal, it could be radial or circumferential; it could be
simple, cylindrical fins. Along with that we can also understand, that the profiles
could be rectangular, it could be trapezoidal, it could be truncated conical, it could be
parabolic. That means what, that the fin surface area can be changing along the length of the
fins. So, we will say, that this is the length of which one is protruding outward, that is,
the length of the fin, this is the length of the fin, this is the length of fin, so
that surface area is changing along the length of the fin.
So, one thing we can understand from here. So, if there is a heat transfer fin here into
this direction, so what is happening here? The heat transfer area remains constant all
through, but here the heat transfer area is continuously firing here. Similarly, the case
here, heat transfer is continuously firing. So, these are the situations, which makes
the system little bit more complicated when there is a variation in area.
Now, we will see that through some mathematical derivations we will try to find out how the
temperature varies along the direction of the fin, along the direction of the fins how
the temperature varies. And that is the very important information that we have to find
out to, find out, that how much heat transfer is possible by the fin? How much enhancement
in the heat transfer rate is possible by the application of fins? If you have to now know
that, we have to do some kind of mathematical analysis for that.
So, we will start now some mathematical analysis here and see that how the temperature profile
for the fins are possible. So, to start with what we will do is, we will see a rectangular
fin and for that let me draw the diagram first. Yeah, so, this is the fin, that is extended
to this direction and... No audio 11:40 to 12:06
So, consider it is a, this is a surface and this is, so this is the fin, that is the rectangular
fin and that is protruded and say, the, the length, that is, it has protruded out, is
say L is the length. And say, the, this is the thickness, if I denote it by H, and so
this is the breadth of the fin, if I denote it by say capital B. And say, this is the
direction of heat transfer, that is, x and consider, at certain point x consider a differential
element of thickness dx. Now, what I was trying to say is, that this
is a solid wall kind of thing and there is a fin, rectangular fin, that is protruding
out and this wall is at some hot, you can say it is a hot surface, so it is a hot surface.
So, temperature, heat is being transferred from the surface through the fins to the environment.
And say, environmental temperature is, T infinity is a temperature at the environment and temperature
of the surface is say T, we will say, that temperature at the surface we will take as
T w and wall, wall temperature. And we can see that the heat transfer area is basically
this area, this area is the heat transfer area from this, as well as this will be the
heat transfer area, from this surface and this surface.
Now, from the hot surface the heat transfer area is this one, this is the heat transfer
area, which is nothing, but this area, this is heat transfer area. So, from this, this
area or this, through this area, this I, because it is a rectangular, this is a rectangular,
separate one, this is the same as the rectangle, so the heat transfer here remains all through
the constant. It is, it is coming from the hot surface to this and then, from the fin
it is uptil this point, uptil this point it is coming to both the sides, it is just converting
out in both the sides of the environment and when it comes here, the heat is converted
from this side. So, what is happening when you say about the
differential section, what is happening, that heat is coming here, part of the heat is coming
to this side and part of the heat getting convected away from the surface to the environment.
So, if we just do the mutual balance and we have, we have discussed is a, is a steady
state phenomena and it is one-dimensional phenomena, that means, heat transfer is only
on x direction and it is a steady state, there is no accumulation of heat energy in any section
of the fins or during this process. So, what happens is if you do the energy balance
for the differential section, but if we find is, that rate of, energy in, heat energy in
at left of the differential control volume, left phase, that would be equal to the rate
of heat energy out at right phase plus rate of heat energy convected away to the environment.
So, then, what we find is, rate of heat energy in at left phase would be and we can understand,
that this is a solid body, so heat transfer will by the mode of conduction within the
fins types now, so will be by the mode of conduction.
So, again we will go back to Fourier’s phenomenal law of conduction. What you see, the rate
of heat transfer at this point is minus K, K is the thermal conductivity of the fin material
and area, the area would be, say, if I say this is A, A is the area of heat transfer,
which is nothing, but this heat transfer area. In this case, it is coming like this way,
heat transfer area and it is, and as, because this one-dimensional and steady state, I will
write perfect differential, that dT by d x at point x, that is the incoming energy and
this will be equal to the rate of heat energy out at right phase. That means, minus K into
A into dT by dx at point x plus dx, that is going out plus something else is going out
in the form of convection to the environment. So, what we will write is that it is equal
to, plus h is the convective heat transfer coefficient between or the fin material and
the environment and h into area of heat transfer. What is the area of heat transfer? Area is
the equation, this is the perimeter or this is the perimeter, this, this is your perimeter.
If this is perimeter, so perimeter into the dx, so this is the perimeter into this dx,
that will give you, the, this fins total convection area, that is equal to P into dx. And then,
the temperature, at this point temperature will be, surface temperature is T, I do not
know what is the temperature we are finding out, how the temperature, del T minus, and
environmental temperature is T infinity, is the temperature of the environment.
So, then, this is the basic equation and from there what we can find out, if we do the simplification
of this equation we could get, like d 2 T by d x 2 into dx K into area and that we will
be equal to h into P into dx into T minus T infinity. It says, that d 2 T by dx 2 minus
h into P into p by K into A into T minus T infinity equals to 0. Now, we will define
terminology m is equal to root bar hP by KA. This implies, that we can write this equation
as d 2 T by dx 2 equals 2 or minus m square T minus T infinity equals to 0.
Now, we will define temperature, the reduced temperature is T bar, if you say another thing,
T bar is reduced temperature and that is equal to say, T minus T infinity. Then, we can write
this implies, that d 2 T bar by dx 2 minus m square T bar, that equal to 0 because that
dT bar is equal to dT as T infinity is constant, so this is the equation.
And now, what we can get? We can get a solution of this equation is a simple differential
equation and the solution of this equation is of the form, that you can go, find out
any differential calculus book and you find any mathematics books, we will finding out
the solution of this equation and the solution of the equation is dT bar equals to, is written
like, that C 1 e to the power m x plus C 2, I am sorry, C 1 into the minus m x plus C
2 e to the power plus m x, here x is equal to T bar is equal to this. Also, this solution
can be written in another form taking the, in the hyperbolic form this is also written
as, say A 1 cos hyperbolic m x plus A 2 sine hyperbolic m x.
Now, we have got the expression for temperature with some constant C 1, C 2 or A 1, A 2, as
you like the way you want to put the format, you want to put. Now, we have to find out
the values of this constant.
So, to find out the values of the constant, as we, as the differential equation is a secondary
differential equation, we need to have two boundary conditions for this. So, the boundary
conditions are, the 1st boundary conditions is pretty simple. It says, that at x equal
to 0, so T bar will be equal to T w minus T infinity and that let us say, that this
is equal to T naught bar, at x equal to 0, it is say, T naught bar.
Now, what about the 2nd boundary condition? Now, there are certain aspects in regarding
the second boundary conditions. Second boundary condition means, it is the values of, it is
value of will certain length of the fin. So, what we will do is that we will define it
into three specific cases. If I say the case one, case one, we will say that the fin is
infinitely long, what does it mean? If a fin is infinitely long, so that we can understand,
that temperature of the fin, if we just say, that this is the length of the fin, so it
was, it will be gradually changing, changing, changing and, and it will be asymptotic to
be and T infinity values. So, this is the T infinity value and this
is what I say, T w at the wall, so it will, finally it will become asymptotic. So, if
you at this point, at this point T w, also temperature, sorry, temperature would be nearly
equal to T infinity at this point. So, and this is happening when one tends to infinity.
So, this is the condition, that fin is infinitely long. When the fin is infinitely long, then
we can say, that x tends to infinity. We will have T bar is equal to T infinity minus T
infinity and that is equal to 0. So, this is the 2nd, that is, secondary boundary condition
for case one, and first boundary condition is known to us.
Now, for case two this is not also the situation and practically, this is never a situation
because as we, what is the use of this particular section? When the temperature is equal to
the environmental temperature, there will no further exchange of fin between the fin
at the environment. So, this is never a practical situation, but this is a theoretical case,
this hypothetical case, this A 1 particular condition, we can, which can be existing.
Now, in case of case two, this is the most practical situation when you have, that fin
has got a definite length. So, fin has got a, fin is of definite length. So, if the fin
is say, suppose the, it is, it is the case one, there case, case one, and in case two
the fin has got a definite length, say suppose this, this is the length of the fin, the definite
length. So, under that situation what happens is, that under as, as it, say steady state
situation, at this point, at this steel, at the end whatever is coming out by conduction
has to be going through these by convection. What I was saying is if you see here, that
is a definite length L, the K condition, second condition under this situation, it is coming,
coming, coming and from this cycle, from this side is going away to the environment, but
some, it is coming here. And form this region whatever is come to by conduction has to go
out to the environment by the mode of convection. So, that is the boundary condition that you
should apply here. So, if that is the boundary condition, then
what we will say is that heat is coming by conduction uptil point n, at point L is minus
K into area into dT by dx at x is equal to L, should be equal to again, the H into area
of heat transfer into T minus T infinity. So, at point x equal to L, so what does it
mean, that minus K dT by dx at x equal to L should be equal to h into T minus T infinity
at x is equal to L. So, this indicates, that this T will be the temperature at x equal
to L. So, this is the 2nd case or 2nd boundary condition.
And then, third case for the third boundary condition is, the tip of the fin is insulated
tip or you can say, or end of the fin is insulated, what does it mean? That means, that if you
go back, if you see here, that this part is insulated, so that is no heat transfer from
this side to the environment. So, if you consider the case two, in the case two situation what
you have written, that minus K dT by dx at x equal to l is equal to this. So, this does
not exist because it is insulated, so this will become 0. So, that means, what I will
write is minus K dT by dx at say, x equal to L equals to 0, that implies, that dT by
dx at x is equal to L is equal to 0. So, this is the condition for the 3rd case and this
is the 2nd boundary condition for the case three.
Now, we have understood that three different conditions can prevail in case of fin. And
also, we have understood that what should be the boundary conditions for different cases
that we have seen. Now, if we try to apply these boundary conditions to solve the basic
differential equation, we will find out that what happens to be the temperature profiles
for different cases. Now, let us start with case one to find out the temperature profile.
So, now we are finding out the temperature profiles. So, for case one and we know the
boundary conditions, now for case one the boundary condition if I summarize here is
that three things we know, one is the temperature profile we know, general temperature profile
is T 1 is equal to C 1 e to the power minus m x plus C 2 e to the power plus m x. This
is the general temperature profile and the boundary conditions with it is that T bar
at x equal to 0 is equal to T naught, but and we have seen, that for the 1st boundary
condition, the 2nd boundary condition for case one is at x tends to 0, T bar equal to
0; as x tends to 0 for infinity, T bar is equal to 0, that is what.
So, now, if we apply these two boundary conditions what we find is that due to application of
first boundary condition we get T bar 0 is equal to C 1 plus C 2. And 2nd boundary condition
says, form the second boundary condition we can find out, that 0 is equal to 0 plus C
2 e to the power m x and this implies, that C 2 is to the 0. Then, what you find out,
then we can write the T naught bar is equal to C 1.
Then, we finally, get the expression as, that T bar by T naught bar is equal to e to the
power minus m x. And we know, that T naught T bar is equal to T minus T infinity by T
naught bar is equal to T w minus T infinity. So, this is the temperature profile for the
first boundary condition. Now, the temperature profile for the second
boundary condition, it is little complicated. So, let us see, that for second case, second
case, case two we have, now here we will write in the temperature profile, for the simplicity
of the analysis we will write the temperature profile as like this, where T bar is equal
to A 1 cos hyperbolic m x plus A 2 sine hyperbolic m x, this is the temperature profile. And
the boundary conditions for these are, as we told, that T bar at x is equal to 0. Already,
we have seen, is equal to T bar 0 and the other boundary conditions we have seen, that
this one minus K dT by dx at x is equal to l minus K dT bar by dx at x is equal to L
will be equal to h into T minus T infinity at, is equal to L. So, this is the second
boundary condition. Now, we have to apply these boundary conditions
over here. Once we apply the boundary conditions we will see, that from first boundary condition,
so this is first boundary condition, this is the second boundary condition, from first
boundary condition we get T naught bar is equal to A 1. If you put x is equal to 0 cos
hyperbolic m x is equal 1, so this one and sine hyperbolic equal to 0. So, T naught bar
equal to A 1, this is the boundary, first boundary condition we get this.
Now, we will apply second boundary condition. For second boundary condition what you have
to do? You have to find out the dT bar by dx and you have to find out at x equal to
l, so dT bar by dx at x is equal to L is minus K A 1 m sine hyperbolic m L plus A 2 m cos
hyperbolic m L and minus K dT by d x equal to this, and this is equal to h into T minus T infinity at x
is equal to L, and that is equal to h into A 1 cos hyperbolic m L plus A 2 sine hyperbolic
mL.
Now, from these two, just simplification of this equation we can do like this way. So,
we will write, A 1 into sine hyperbolic mL plus h by m K cos hyperbolic m L, that is
equal to minus A 2 cos hyperbolic mL plus h by m K sine hyperbolic mL. Now, using this
we can find out, that A 2 equals to minus A 1 into sine hyperbolic mL plus h by m K
cos hyperbolic m L divided by cos hyperbolic mL plus h by m K sine hyperbolic m L. So,
then, we know what is A 1, we know that, and if you put expression over here, then we can
get A 2 is equal to minus T naught bar into sine hyperbolic mL plus h by mK cos hyperbolic
mL divided by cos hyperbolic mL plus h by m K sine hyperbolic mL. So, this is becoming
the expression for m A 2.
Now, we will find, we will find out for T. So, T bar equals to or T bar by T naught bar,
if I write directly, it is becoming. Now, if you, the values of A 1 and A 2, we will
get an expression like this, cos hyperbolic mx into cos hyperbolic mL plus h by mK cos
hyperbolic mx sine hyperbolic mL, then minus sine hyperbolic mx sine hyperbolic mL minus
h by mK sine hyperbolic mx into cos hyperbolic mL and this is divided by denominators, we
have cos hyperbolic mL cos hyperbolic mL plus h by mK sine hyperbolic mL. Now, from this
we can write, that T bar by T naught bar is equal to T minus T infinity by T w minus T
infinity and this is equal to, you see, that cos A, cos A and cos B minus sine A sine B
gives cos hyperbolic A minus B. So, is basically cos hyperbolic m into L minus x is m, m equal
to minus x plus h by mK into sine hyperbolic m into L minus x, divided by cos hyperbolic
mL plus h by mK sine hyperbolic mL. So, we can see that, that is pretty similarity in
the expression and this is becoming the temperature profile for case two; this is the temperature
profile for case two.
And now, if we see for case three, for case three, again if we write the temperature expression,
that T bar, already we have seen generalized temperature expression is A 1 cos hyperbolic
mx plus A 2 sine hyperbolic mx. And the boundary conditions are, we know, that T naught T at
x is equal to 0, reduced temperature is equal to T naught, this and this is the, I will
say, that first boundary condition and the second boundary condition is, you know, that
dT bar by dx at x is equal to L and that is equal to 0. This is the secondary boundary
condition. If you apply the boundary conditions over here, from the first boundary conditions
already we have seen that and it generates, that T bar 0 is equal to A 1 because this
part is 0. And from the 2nd boundary condition we can write, that T bar by dx at x is equal
to L equal to 0 equals to A 1 m sine hyperbolic m L plus A 2 m cos hyperbolic mL.
Now, from this you can write, this implies, that A 2 equals to minus A 1 tan hyperbolic
mL. So, from this we can write, that T bar equals to A 1. If you put this expression
over here, we will write, we will get cos hyperbolic mx m to cos hyperbolic mL minus
sine hyperbolic mx plus sine, sorry, hyperbolic mx into sine hyperbolic mL and this divided
by cos hyperbolic mL. So, that means, this says we know, that A 1 is equal to T naught
bar, we can write, that T bar by T naught bar again, is equal to T minus infinity by
T w minus T infinity and this is equal to, this is cos A cos B minus sine A sine B, so
we will write cos of L minus, sorry, cos hyperbolic mL minus x by cos hyperbolic mL. So, this
becomes the temperature profile for case three when the tip of the fin is insulated, that
is why it is tough part from to h. Now, we have already got the three temperature
profiles in the fin, which is a function of the values x, the length of the fin, that
we have seen and we have seen for three different cases where the boundary conditions are different.
Now, our next target is to see, that how efficient the fins are because what we have understand,
if we see, if we go back to the fin diagram, if you try to understand, that as we go along
this direction, temperature will, is expect, expected to be, as we go in this direction
temperature is expected to be decreasing; temperature is expected to be decreasing along
with them. And as temperature is decreasing, the driving force of heat transfer from the
surface of the fin to the environment, that is, T minus T infinity will decrease and if
the driving force decreases, the efficiency the heat transfer will be reduced.
So, efficiency of the fin come, because of the fin you would, will be that much efficient,
which had been going. It was the initial stages of the fin where the length is small, here
the fin will be more efficient, then this part because in this part the driving force
is less though the area is same. The heat transfer from this part will be much less
compared to this region. So, greater the heat transfer, rate will be greater decrease from
the fin surface. And therefore, we need to find out, that how long we should put the
fin, how far we should put the fin? If you put for infinity long fin, what happens to
the efficiency and if you go for a small fin, what happens to the efficiency? So, we have
to find out expression for efficiency of the, of fin for which the heat transfer is taking
place from the fin surfaces. And the one thing we should understand, that
before we go to discuss our efficiency one thing we should understand, that whatever
a steady state, whatever energy has come through this surface at this point, whatever energy
has come to this fin at this point, has to be transported by the fins because this is
study state condition. So, heat coming at this point will be transported by the fins
at different places. So, therefore, we should say, that heat that has entered into the fin
at this base is nothing, but the rate of heat, that has entered into the fin is nothing,
but the actually transfer by the fin. So, and what will be the theoretical or ideal
condition? Ideal condition would have been that in all the surfaces, say of the fin,
would, would have been maintained as this temperature, wall temperature is the higher
temperature. And everywhere, that profile or the gradient would have been same if every
hour the gradient is same when heat transfer, the driving forces remain the same. That means,
as I was telling, that fin is becoming lesser percent as we make it more and more longer,
correct. Now, this is becoming, because as it, as it is becoming longer the temperature
is gradually decreasing, so the driving force, driving force between the environment and
the fin surface is decreasing. Now, if you can maintain the same driving
force, that means, if you can maintain the temperature of the fin surface all through
same, then what is the performance of the fin at this point will be, the same performance
would be occurred by the fin at this point. But that is not practically possible because
there will be always a temperature drop from this side to this side and therefore, then
efficiency term comes. So, when we say about this part, which is not practically possible,
that is for ideal situation and this is the ideal situation or ideal condition that the
whole surface of the fin is at base temperature; that is called ideal condition.
Therefore, what we write is, that efficiency of the fin is written as, defined at eta a,
which is nothing, but fin efficiency and this is equal to actual heat transferred divided
by heat would be transferred, this a hypothetical case, ideal situation if entire fin area were
maintained at base temperature. Base temperature means, in this present situation it is nothing,
but T w, wall temperature, that is the base temperature, but this is an ideal situation,
this never a practical situation and this is the practical situation. Now, this can
be, we can write it as, that actual heat transfer and that is equal to, as I told you, that
it is nothing, but minus K into area into dT by dx at x equal to 0, and so this is equal
to, that we can find out. Depending upon the temperature profile for
different cases we get, we will find out form this, that minus KA dt by dx at x equal to
0, we have found out the different temperature profile for different situations. So, with
this expression we can find out what would be the actual heat transfer and ideal condition
is pretty simple, that ideal heat transfer will be ideal in this situation.
It is, say we can say, that h into the total area, if I see this is the total area of heat
transfer, total area of heat transfer hA T, total area of heat transfer into T minus T
infinity, that is nothing, but you write as this as, hA into and that this T will be T
w because everything is maintained at T w. So, hA T into T o bar. Therefore, fin efficiency
we can write as minus K into A into dT bar by dx at x equal to 0 because you know, that
dT is equal to d bar divided by h into A. This is total heat transfer area from the
fin, that is useful for heat transfer by T naught bar.
So, using this expression we can find out the fin efficiency of the fin, that is the
used and there is another term that is called fin effectiveness of the fin effectiveness
of the fin. This is given as, that rate of heat transfer with fin divided by rate of
heat transfer without fin, sorry, rate of, without fin. So, this is also one important
thing, that we would like to see, that for any particular system what is the effectiveness
of heat transfer.
So, we can find out, that efficiency of the fin, effectiveness of the fin, we will see
for different case of, we have discussed what is happening to the effectiveness, efficiency
of the fin, that in the next lecture we will discuss on, that how is the efficiency of
the fin for different cases and that we will take up a problem. And to see, that how to
calculate, that fin efficiency, as well as, heat transfer, total heat transfer due to
fin surface and as well as the effectiveness of the fin, we will discuss in the next lecture.
Thank you very much.