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Welcome, to lecture three of module two. In last few lectures, we have discussed about
heat transfer by conduction only, in cartesian coordinate system and cylindrical and in spherical
coordinate system, but in practice if you see that heat while heat transfer is involved
there are some fluid is also involved and the solid walls or say cylindrical walls or
spherical walls they do separate some fluid the fluid can be either under stationary condition
or can be under flowing conditions, and the heat transfer in many situations happens to
be from a hot fluid to the cold fluid through the barrier of the solid barrier while the
conduction takes place. So, we will in this lecture we will start seeing that when there
is a combination of convection and conduction how the heat transfer phenomena takes place.
Already we have discussed little bit on basics of convection here we will be discussing a
little more on this particular phenomena.
Now, as I told you that a solid wall consider a solid wall, and this is separating a 2 fluids
and say this one is the film in this left side, and this one is film in the right-hand
side, and this side is say hot fluid and this side is the cold fluid. So, we have a solid
wall and this is the solid wall, this is the solid wall there is a hot fluid there is a
cold fluid and this hot fluid or cold fluid can be moving fluid, can be stationary fluids
and this solid wall is separating the 2 fluids. Now what happens is heat transfer takes place
from the hot fluid to the cold fluid through these solid wall and we have seen that this
kind of examples, are in most of the situations we have like, heat exchangers you have other
pipe walls are acting as the barrier between the hot fluid and the cold fluid. If we see
in place of a refrigerator the the wall of the refrigerator acts as a barrier between
2 almost stationary fluids the fluid inside the refrigerator and the fluid outside the
refrigerator; that means, home environment that is in many situation it can be viewed
as a stationary fluid. So, like that you can have both stationary
fluid or you can have some flowing fluid where the solid is acting as a barrier and heat
transfer phenomena can be encountered in all the cases. Now if we try to see that whether
it is a stationary fluid or a solid or a flowing fluid in all the cases there is expected to
be some kind of flow finally, being developed in either sides of this and in case of stationary
fluid we know already we have discussed the flow development of the fluid element takes
place due to the changes in the density and due to the and that happens due to change
in temperature. So, there is a circulation that has been generated and in case of flowing
fluid naturally the circulation is always there due to application of the external energy.
So, in either of these situations there is there will be some film in both sides of the
walls, you say it is the hypothetical or imaginary film that exists the idea of this film is
that say, if we see that the in the hot fluid this is the temperature and this temperature
profile will be changing like this and then temperature profile in the solid medium also
will be changing like this and temperature profile in this medium also will be changing
like this, and it will go to this side so; that means, what so this diagram is a temperature
profile and we can see that if we say that the hot fluid side is 1 and cold fluid side
is 2. So, we can say that say this is T 1 bulk temperature of the bulk here, we can
see the temperature is constant; that means, in this bulk the temperature is constant and
the fluid is well mixed and there is a constant temperature and there is the highest temperature
hot temperature in the system. And then here it says at this interface the surface and
this fluid interface that the temperature is say T 1 and then here again say in the
interface T 2 and here it is again say T2 b. So, what is happening is you can see that
there is a this profile and this profile we have I have drawn it to be a little non-linear
in nature whereas, that the profile I have drawn it to be in the in the solid linear
most of the cases in the solid if we have discussed if it is a isotropic material and
if the thermal connectivity does not change then and there is no heat generation term
then this within the solid the temperature profile will be linear, that we have already
told and here we have written that this2 are non-linear temperature profiles. Now if we
see if we minutely observe that the very closed vicinity of the wall the very close vicinity
of the wall the temperature profile remains more less abstract line, but afterwards if
we move like this afterwards that it is it is a non-linear.
So, initial period it is a stationary it is a linear 1 and then afterward it is a non-linear.
The reason behind these kind of thing is, what is happening in this case that fluid
layers as we have discussed in the previous cases also though there is a circulation or
movement of the fluids the fluid adjacent into this layer has got a no sleep condition
and, because of the no sleep condition the fluids are sticking to this and there will
be a conduction.
And gradually that the fluid will get some movement there will be some kind of mixing
and there will be some kind of a convection heat transfer. So, conduction to convection
there will be a transfer or transition from purely conduction phenomena to the purely
I should not say purely, because in the convection there can be some conduction also. So purely
conduction phenomena to and after civil amount of convection phenomena. So, there is a gradual
transaction takes place transition takes place and therefore, you will be having some non-linear
profiles over here; however, if we say that from electrical analog if we write the diagram
we will find that there are 3 resistances now that we can very easily understand. So,
this is the thermal potential T 1 b and here is the resistances r 1 this is thermal potential
T 1 and this is thermal potential t 2 and this is the thermal potential T 2 b and here
resistance r2 and this resistance is r3. And what is this r 1 already we have seen
r 1 is the convective heat transfer resistance and this I will denote as r 1 as 1 by h 1
into area, we will assume that it is a flat uh this is flat and the area is constant.
So, it is for a plain wall we consider and. So, area remains constant area is no changing,
but in case of cylindrical coordinate and in case of spherical coordinates the this
area and this area are going to be different that already we have seen and we will be seeing
further also. So, r 1 is this and r2 is equal to say if this thickness is l then it will
be l by k a and r3 is again 1 by h2 into a where you can understand that h 1 and h2.
So, h 1 and h2 these are heat transfer coefficient. So, convective heat transfer coefficients
at 2 different locations in the station side 1 and station side 2.
And we know that this we have already discussed the values of h or convective heat transfer
coefficients these are functions of h is function of as I told you fluid property flow property and geometry. So, if we see
that we will write that is a function of in the fluid property we have viscosity sorry
new density rho specific heat k c p thermal conductivity k similarly, we have flow property
velocity and geometry say characteristic length l l is called characteristic length and this
l is equal to diameter l is equal to diameter for say tube l can be length for flat plates.
So, depending upon the flow situation basically this l will be discussed. So, we will be seeing
in later in future when we will be discussing more on convective heat transfer there will
be finding out how h is a functional characteristic length. Now if we try to analyze some more
thing like once we have understand the r 1 is equal to this r2 is equal to this r3 is
equal to this, and under steady state condition q dot would be equal to we can say that heat
transfer is taking place from here to here from this side to side then this side and
then this side. So, I will write we assume that heat transfer
in the positive this direction. So, we will write q dot is equal to h 1 into a into t
1 b minus t 1. So, this is a positive quantity and this is also equal to t 1 minus t2 by
l by k a and this is also again equal to h to a into t2 minus t2 b this is the driving
force. Now we have seen that Q dot is equal to that this is the convective transport this
is the conductive transport and this is again the convective transport.
Now, in case of some more understanding about this film if we try to say that suppose there
is a film that as I told you there is the imaginary field and there is a gradient of
temperature all through and that is a non-linear temperature profile means, that the gradient
is continuously changing. Now if we consider that this film consider this film has hypothetically
we will see that in the next week, if you see say suppose this is the thickness and
1 side I am drawing. So, this is the this is the wall and here is the profile and if
we say that if I tell that this 1 is delta that is the film thickness and in case if
we assume that in the film if we assume it to be a stationary film and if we assume that
heat transfer by conduction. Then what we get is in this case we can write that say
q dot rate of heat transfer according to the conduction more it will be k into area say
minus into say T 1 minus T 1 b this is T 1 b and this is T 1 and by delta.
And we have seen that this is also if you write is equal to h into area into T 1 minus
T 1 b by delta. So, take out this is not necessary. So, under these conditions. So, if we consider
that heat transfer is both at this direction. So, either it is negative both will be negative.
So, heat transfer will be in this direction. So, under this situation what we can find
is i am sorry this is not there by convective mode it is like this and this is by conductive
mode. So, we can say that h is equal to k by delta; that means, if we assume that conductive
heat transfer if we assume that it is a linear this is T 1 minus T 1 b by delta. So, we have
taken this is a gradient. So, this is thermal gradient between this point in place if we
assume that it is a straight line rather than it a curved 1 then this assumption is much
better. So, if we assume that it is a straight line not the non-linear 1 curved 1 in that
case t 1 minus t 1 b by delta that becomes the driving force for gradient gradient and
minus k a that becomes a total heat transfer rate. Whereas, if we show from this point
from if it is a convective 1 if you discuss this from convective angle then it will become
minus h a into t minus T 1 b so; that means, what we say is here that h is equivalent to
k by delta so; that means, when that heat transfer phenomena is by conduction on this
stagnant film and then h can be written as k by delta; that means, that means h also
has got some kind of linkage with the conductivity and therefore, sometimes h is also called
film conductance as already we have discussed it is also called film conductance.
But, what it when we find that h is equal to k by delta we assume that whole thing is
stationary whole thing is whole thing in this is stationary. The film is stationary and
heat transfer is solely by conduction in that case h is equal to k by delta. So, but in
practice in practice it is never the situation in practice therefore, we get a non-linear
1 the temperature profile we can see that the temperature profile remains linear very
close or very almost to the vicinity of this wall while it is almost by conduction mode,
but in the other when distance apart from the wall it is not only by conduction mode
there is some movement of the molecules are taking place and that is being governed by
the hydrodynamics. That has been governed by the hydrodynamics of the fluid in this
region. So, so and the hydrodynamics is again characterized
by the fluid properties by the flow properties and the geometrical configurations and therefore,
therefore, when we say that h h is basically the convective heat transfer coefficient is
the function of all these 3 properties as I have discussed. And we should not what I
meant to say here though theoretically we can see when there is a perfectly stationary
field that, we can find out h is equal to k by delta by in practice h is not calculated
based on this equation, because in practice we do not know exact the delta is is not that
the field thickness power when the fluid is stationary. So, the field thickness takes
care of in practice whatever I have drawn in the previous figure the field thickness
takes care of some movement of the un fluid elements over here and therefore, the age
values are calculated with the help of experiments and there are lot of theories lot of empirical
relations are there to find out the values of h that we will discuss later on.
Now, if we consider a case which is a very common case say for a heat exchanger, what
is there in a heat exchanger heat exchanger, you have inner fluid that is flowing through
in a tube in a tube this is the central line. So, this heat exchange you can consider is
a double pipe heat exchanger let us for the time being assume that and this is the thickness
of the tube wall. And consider this is the outer tube, and the idea is that there is
a fluid which may be flowing in this direction, and there is a fluid say flowing in this direction
which is the countercurrent flow, but whatever may be the situation what happens is say this
is equal to your r i and this is r o o this happens due to the thickness of the fluid
and then say say suppose this is your hot fluid and this is a cold fluid. So, heat is
being transported from the hot fluid to the hot fluid to the cold fluid. So, heat transfer
is taking place in this direction ok
Now, how many resistances we countered as we have seen in the previous cases that were
a flat plate that was a flat plate plain wall now here these areas are changing. So, what
we will write is here, again if you write the analog electrical analog then you have
again there are 3 resistances and the very fast resistance the inner resistance is again
this is 1 by a i into h i a i is the area of the surface area of the tube of the inner
side is based on the r i and here is the resistance of this tube wall and this is given by l n
r o that already you have discussed in few last classes twice pi k into l l is the length
of the l is the length of the say pipe system round pipe system and here it is 1 by a o
h o. So, this way we can see that q dot we know that is equal to T i minus t o with the
temperature here, it is T i and you write is t o the hot fluid temperature is T i cold
fluid temperature is t o. Then this will be the summation of the all the resistances.
So, then what we can say there is there is a concept that is call overall heat transfer
coefficient or overall heat transfer resistance. So, overall heat transfer coefficient is found
out overall heat transfer coefficient is found out from the overall heat transfer resistance
and overall heat transfer resistance is the summation of the total resistances. So, so;
that means, overall heat transfer coefficient is found out from the r t inverse of the total
resistance or overall resistance which is found out from the inverse of the summations
of r. So, that way if we see that 1 by u i a i is
equal to that i am sorry we have told r t is equal to summation of r yes 1 by r t is
equal to your fine. So, this is the total resistance this is total resistance is the
summation of all the resistances if we go by that. So, 1 by u i a i is the total resistance,
and why the u i a i or rather u i is the overall heat transfer coefficient based on based on
a i or inside area. So, u i is the overall heat transfer coefficient and 1 by u i a i
is the total resistance based on the inside area and this is equal to 1 by a i h i plus
l n r o by r i by twice pi k l plus 1 by a o h o. Now if we do little simplification
you can find out this implies that 1 by u i is equal to 1 by h i if i multiply by area
plus a o i am sorry we will write it as r i a i is equal to again twice pi r i into
l. So,2 pi l2 pi l cancels r i l n r o by r i
by k plus r i by r o into 1 by h o. So, this is becoming this is becoming the expression for overall
heat transfer coefficient and is the inverse of the overall heat transfer coefficient.
And this is based on the inner area similarly, we can have the expression for 1 by u o this
is based on external area and that is equal to r o by r i into 1 by h i plus r o l n r
o by r i by k plus 1 by h o. And incidentally we can understand that 1 by u i a i is equal
to 1 by u o or 1 by u o a o or u i a i is equal to u o a o so product of so, we can
we are defining here 2 overall heat transfer coefficients u i and u o both u i and u o
they are overall heat transfer coefficients, but varying their values of different because
they are based on different areas, but the product of the overall heat transfer coefficient
u i into a i is equal to the product of u o into a o. So, we can understand a equals
to a o is more than a i. So, u i will be lower than u o i am sorry
as because a o is more than a i. So, u will be lower than u i now we will try to discuss
some important aspect that is called critical insulation thickness that is being encountered
this kind of aspect is encountered. So, this is what I am saying is the critical insulation
thickness. This phenomenon is encountered particularly when you have the heat transfer
devices either in cylindrical mode or in spherical mode or spherical system. If you say that
whether in case of cylindrical coordinate or in or in case of cylinder or in case of
sphere, if we go on increasing the insulation if you go on increasing the insulation thickness
what happens the external surface area goes on increasing, after some time what is happening
the thickness of the insulation is increasing. So, the thickness of the insulation that acts
as a resistance for the heat transfer whereas, area of the insulation it acts just against
to this if in area increases as we know that from the external surface that heat transfer
rate is u is equal to h a delta t. So, a increasing in area means the heat transfer
rate increases, after some time when your delta x changes we have seen in case of conduction
in case of conduction the resistance is delta x by k into area and it case of convection
the resistance is 1 by h a. So, what we can say that if you consider a case of say cylinder
or a case of a sphere, as we go on increasing this is to say for a cylindrical it is say
top view of the cylinder or you can see that or if the cylinder is lying this way horizontally
it is the front view of the cylinder. So, what is happening if you this is to say in
the cylinder?
So, this is the thickness of the cylindrical pipe and this is r i and this is the r o.
So, what is happening if you go on increasing the thickness? So, delta x is increasing and
delta x increasing means, resistance is increasing. So, heat transfer rates should be decreased
resistance increase means, resistance of heat transfer rate should be decreased, but at
the same time if you say that from this side to this side the convective transfer is taking
place and as you go and increasing this. So, this area is getting increased and as, because
the area is increased. So, heat transfer rate is again further increasing. So, there should
be a an optimum situation that will decide that whether we should or which is called
also that critical insulation thickness that will decide the insulation thickness particularly
for the cylindrical devices or spherical devises. Where that with increasing the insulation
thickness the area changes, but in case of plain wall this kind of concept does not exist,
because there is no change in the area if you go on increasing the insulation thickness.
Now we will take off this case and we will see that what happens what is the insulation
thickness critical insulation thickness for spherical and for cylindrical coordinates
now. So, critical insulation thickness now we will consider in the both the cases it
will look like this similar, and what we will do this in this case we will assume that the
thickness of the pi or the spherical vessel they do not play much significant role particularly
in the heat transfer rate, because we have seen that the that the that the thickness
of the pi or thickness of the cylindrical plate that the percentage contribution to
the total resistance is almost negligible. So, we will assume that thickness is more
participating in the in controlling the heat transfer rate. So, that is the major assumptions
now what is happening is say. So, say this is r i and so this is say r o. So, it will
look similar in case of cylindrical and the spherical system only thing is that the surface
area as an all these things we will take care accordingly it will be different. Now what
assumptions you are taking that we assume that pi or a sphere thickness resistance offered
by this is offered by are negligible, and we will consider only 1 kind of insulation
we will consider. So, under that situation for if we separately write for cylindrical
and spherical cases. So, what happens is we can write the expression for Q dot r already
we have seen that is this happening in radial direction there are 3 resistances again that
inside resistance due to this insulation and the outside resistance there are 3 resistances
we know. So, if you write then Q dot r will be equal to T i minus say T infinity.
So, this is the T i is the temperature of fluid here and T infinity is the temperature
of the fluid here that is the environment temperature here it is this. So, from t i
is the temperature of the bulk fluid from T i there will be a convective resistance
then it is coming to the surface there is a convective resistance and then again there
is a convective resistance and then coming to the environment. So, this will be T i by
3 resistances already we know that is 1 by h i2 pi r i into l l is the length of the
cylinder plus l n r naught by r i by twice pi k into l k is a thermal conductivity of
this material and then we know plus 1 by h o twice pi r o into l.
So, h i and h o are the heat transfer coefficient the inside and outside region. Similarly,
for spherical coordinate we will have q dot r is equal to say T i minus T infinity by
we will have 1 by four pi h i into four pi r i square this is the area, plus you will
have 1 by r i minus 1 by r o by four pi k plus, we have another area 1 by h o four pi
into r o square these are the 3 area 3 resistances we found. Now, what we have to do if we have
to find out we are interested about not about the inner part, because inner part were not
changing. So, that that r i will remain r i and that is that is not going to change
only thing is that due to addition of insulation r o is going to be changed and, because is
r o is going to be changed. So, we will differentiate q dot r with respect to r o.
So, if we do that then what we will get the q dot r by d r o if you differentiate this
with respect to this and then if we equate it to 0 then we will get a condition of maximization
or minimization we will get a condition of maximization or minimization and then in both
the cases after doing that if we see that d2 q dot r by d r naught square and that should
be that can be positive or negative in the present case this is this will come as negative
and therefore, what we say that d q by d q r dot by d r o is a maximization case maximization
condition. So, that anybody can do that derivation is a just simple differentiation. If we do
the differentiation and if you just make if you do the differentiation of this 2 equations
separately and make it them equated to 0.
And then finally, if you do the second of the derivate then you will find that right-hand
side will become if will be a negative quantity so it is a. So, so the first law derivative
is the maximization condition and the first law derivate when we equate it to 0 then in
case of cylindrical what we will get is the relationship says that r naught will be equal to k by h naught, and in case of
spherical system will get r naught is equal to twice k by h naught. So; that means, that
means that we can get the critical insulation thickness which will give the maximum heat
loss. So, if there is a steam pipe or if there is a storage of vessel cylindrical storage
vessel, and you want to store it at a very low temperature some fluid particularly the
cryogenics materials are stored at low temperatures. So, under that situation we have to say that
we are doing the insulations for the cylindrical or the spherical pipes, you are doing the
insulations beyond this maximization condition; that means the insulation thickness should
be more much more than these critical values. So, that a heat transfer law heat transfer
rate or heat loss is minimized or becomes lower at the same time when you consider a
case of say electrical wire and we have discussed that electrical copper wire there is a heat
generation in case of copper wire. And what we will do we do not want to heat we do not
want that generated in the copper are to be stored within the wire then there can be a
problem. So, in that case what you will say that that copper wire which is the electrical
wire the kind of heat generated will see that let it be dissipated, to the atmosphere let
it to be transferred to the atmosphere in that case, we will see that in those kind
of situations the insulation lies very close to this critical value. And there can be certain
situations when we may find that the insulation thickness in insulation thickness is becoming
lower than r o is becoming nearly equal to or I should say actually insulation thickness
is r o minus r i that we should be very careful, that this is this is the this is the radius
and this r o minus r i is the insulation thickness r o is the outer radius, but r o minus r i
is the actual value of the insulation thickness.
And it may. So, happen that r o minus or critical insulation thickness becomes lower than less
than equal to r i. You can get we can get this kind of situations less than equal to
r i that is the possibility when when the values of k thermal conductivity is very low,
then we may not require actually having any such kind of insulations. So, the problem
is in that kind of situations we cannot find out any such value of critical insulation
thickness, because the critical insulation thickness becomes almost 0 or its negative
that is not practically possible. So, we can come up with this kind 3 kind of situations
in 1 case what we required to put some insulation thickness particularly, is to save any heat
in leak or heat loss from a system under that situation we have to put the insulation thickness
more than the critical value that you have got, in another case when as I told you in
case of electrical wire there we may be requiring to put insulation thickness in such a way
that that just matches the values of the r naught very close to this.
And the third situation is such that when the k values will be very low or h o is very
particularly the k value is very low that r o is coming such a value, that r o minus
r i is less than r o minus r i is less than equal to 0, r o minus r i is less than equal
to 0 or r o is less than equal r i.
Under that situation this the concept of critical insulation will become meaningless. yes But,
practically this is this is the case a from only from the heat transfer characteristics
point of view, that there should be critical thickness in all, but in practice we have
to see some other aspect that is called that is why there is another formula is called
optimum or economic. In practice this we can have economic or optimum insulation thickness
it is something different from the critical insulation thickness, in case of critical
insulation thickness we are just considering that heat loss is minimized to minimize the
heat loss we are trying to focus on that, but in practice what is the you have to see
the economic point of view, we have to see the cost of the insulation, we have to see
that cost of the heat lost and cost of the insulations and there is will be a trade-off
between this 2 and that is called your economic or optimum insulation thickness we have to
find out. So, typically a typical typically if we show
that if we adjust a rough sketch of this is say cost and here it is say r o minus r i
is the insulation thickness. Then we will find out this is say cost of energy heat energy,
and say this is the cost of say suppose this is c 1 and this is c2 c2 is the cost of insulation.
yes As we go and increasing the thickness the cost of insulation will increased, as
we go on increasing the thickness the energy loss will be minimized as we have seen, that
we should do beyond the critical insulation thickness the insulation should be d1 beyond
the critical insulation thickness. So, if you go on doing beyond then there will be
a certain situation that continuously your cost of energy, because the energy loss will
minimized the cost of energy will be reduced, but we see that at the same time that cost
of insulation will be will be very high there can be a certain situation which will take
care of that the total cost we have to find out. This c t total cost of this 2 and there
may give us some kind of minima that may fall in this or that may fall in some other points
of this. So, we have to find out the minima here in the total cost; that means, we have
to find out c 1, we have to find out we have to find out c2 and we have to find out c t
that is equal to c 1 plus c2. And then what we have to do is we have to
differentiate d c t by d r naught and that we have to equate it to 0 we get the value
of critical or economic or optimum insulation thickness. Now how to get the value of c 1
c 1 is the cost of energy cost of energy is related to the heat rate loss of heat rate
of loss of heat into the cost of energy. So, rate of heat energy loss into the cost of
energy. So, if we say that c 1 we can write as q into c into h q dot q is the heat q is
the rate of heat loss, and c h is the cost of energy per unit amount of heat. So, under
that situation for say this is the this is not c 1 this is cost of I should say cost
of energy is equal to say q dot is the heat loss into c h is the cost of energy heat energy.
Now, what happens to the present worth, present worth of that energy is say c 1 that is we
have to calculate present worth of that energy is say j is equal 1 to n; that means, for
n years the heat exchanger is working n years or that that insulation that heat transfer
device is working for n years. So, n years that heat is being lost. So, this is q dot
into c h by 1 plus i e to the power j where n is the service life in years in years and
i is the fractional l 1 compound interest rate. So, if this is the interest rate. So,
1 plus i to the power j.
And summation of this up to n years that c 1 that gives you c 1 which is c 1 is called
the present worth also it is called present worth of the heat lost. And then present worth
of insulation cost of insulation this is your c2. So, c2 we can get that c2 is equal to
c2 will be equal to the volume of insulation depending upon in case of in case of cylindrical
coordinate it is pi into r square minus r i square into l and into say sum c is the
cost of cost per unit volume and this is the total volume in case of cylinder the total
volume of insulation and c is the cost per volume of insulation. Now, you got c t is
equal to total worth is equal to present worth due to heat and c2 the present worth due to
insulation. So, total worth you got and from there if we find out that d c t by d r naught
and we equate it to 0, then will be we will get a value of r naught minus r i we can calculate
and that gives the optimum insulation conditions. So, this way we can find out. So, we have
seen that there are2 types of insulation thickness termed about or called about 1 is called critical
insulation thickness other is called economic or optimum insulation thickness critical insulation
thickness only focuses only the aspect of heat transfer, but economic or optimum insulation
thickness focuses on both economy as well as the heat transfer or heat loss. So, in
the industrialist or in practical situation it will try to look into the economic insulation
thickness rather than the critical insulation thickness. Now I will take a 1 problem that
problem is not directly related to the insulation thickness it is something related to the conduction
heat transfer in a plain wall. So, the problem is like that the steady state temperature distribution
in a 0.5 meter thick plain wall is t equals to 400 plus 1000 x minus 2000 x square where
t is in degree centigrade and x is measured meter from 1 surface. Thermal conductivity
of the material wall material is 25 watt per meter per Kelvin then first question what
is the maximum temperature, what is the maximum temperature in the wall and what is its location,
question 2 what are the surface temperatures, question c how much are the heat fluxes at
the surfaces. So, this kind of problem when we try to see that in the very first part
we are told that what is the maximum temperature. So, we know the temperature profile t equals
to 400 plus 1000 x minus 2000 x square. So, maximum temperature if I have to find
out if i do the maximum condition we will try to find out d t by d x and that is equal
to 1000 minus 1000 minus 4000 x and that is equal to 0. This implies this implies that
x is equal to 0.25 and this is maximum condition because d t by d x2 if you try to find out
that will become minus 4000. So, this is a maximization condition. So, x is equal to
0.25 it is maximum and we have told that the wall is of 0.5 meter so; that means, maximum
temperature happens at the exactly at the middle and then this is the first part then
second part says what are the surface temperatures. So, t at x equal to 0 this is x equal to 0
and here it is x equal to0 point five. So, t at x equal to 0 will be equal to 400 degree
centigrade and t at x is equal to 0.5 and that is equal to it will becoming 400 into
500 sorry 400 plus 500 minus again 500. So, this has become again four hundred degree
centigrade; that means, we can say that maximum is at 0.25 and other 2 points it is 400 400.
So, this is symmetric profile that we are able to get. So, here it is so maximum temperature
takes place at maximization takes place at this point. So, now, what is this maximum
temperature here it is 400 here it is 400 degree centigrade and the first question is
that where maximum temperature we have found out the location in the first case the maximum
temperature we can find out for this case as t max is equal to we can find out that
400 plus thousand into 0.25 minus sorry minus 2000 into 0.25 square. So, from here we will
get t max as if we do the calculation we will be getting as 515 degree centigrade. So, this
is basically 515 degree centigrade. So, this the maximum temperature.
Now, the third question what is being add is what is being asked is that heat flux so;
that means, we have to find out q dot at x equal to 0 which is nothing, but k into d
T by d x at x equal to 0 and because flux. So, it is per unit area. So, k into d t by
d x at x equal to 0 and we know that k already is being even as 25 into d t by d x at x equal
to 0. We have already calculated d T by d x at x equal to 0 will be on the d T by d
x we can get it is 1000. So, this says 2500 watt per meter square is equal to 25 kilo
watt per meter square. And then q dot at x is equal to 0.5 will be equal to again k into
d T by d x at x equal to 0.5 we know d T by d x we have found out. So, it is k is equal
to twenty-five into we have seen d T by d x we have calculated it will be thousand minus
four thousand x and x equal to 0.5 if we put it then you will be getting minus 2 2500 watt
per meter square interesting. So, we can find out that it is again 25 minus 25 kilowatt
per meter square; that means, exactly we have understood, because the temperature in the
both surfaces are same and both are in the environment. So, the temperature heat loss
or heat fluxes should be same, but the opposite side the opposition side indicates that heat
fluxes or the heat loss are taking place in 2 different directions.
If you see in this case heat losses in this direction and in this case heat loses in the
direction. So, 2 different directions gives you the 2 different values like positive 25
kilowatt other case negative 25 kilowatt. So, like this way we have to see or we have
to try we have to practice several problems in heat exchanger and then only we will be
getting more and more confidence in this subject area. So, we will stop here in the next lecture
we will try to discuss on extended heat transfer thank you