Thermal Entrance Region: Uniform Wall Temperature

Thermal Entrance Region: Uniform Wall Temperature


Hello welcome in the 15th lecture of convective
heat transfer course in this lecture we will be discussing about thermal entrance length
in our last two lectures we have discussed about thermally fully developed region mainly
but here we will be discussing at the beginning of the duct what happens that means when the
temperature profile is developing to get a steady state profile okay. So this thermal entrance region we have already
defined in our 11th lecture so there we have shown that at the beginning how temperature
profile actually gets developed in to a parabolic one okay. So here we will be showing that entrance length
region and that two in this lecture we will be considering uniform wall temperature case
okay. So that means the pipe is actually being raped
up with some heating coil which is maintaining the wall temperature at constant okay. If you have the other extant that means the
heat flux is constant so whatever the heating coils are supplying to the fluid inside the
pipeline that is actually heat flux is constant that case you need to see in the next lecture
okay. So as I have mentioned that we will be discussing
about thermal entrance region over here, thermal entrance region with uniform wall temperature
okay. So let me at the beginning tell you that what
will be the outline what things will be covering in this lecture we will be introducing the
concept of thermal entrance length. In hydro dynamically fully developed but thermally
developing in flow in pipe. This assumption we are doing over here that
we are considering the flow hydro dynamically has developed that means it has taken a parabolic
velocity profile but temperature has not yet grown in to a steady state profile okay. So though it is hydro dynamically developed
fully developed but thermally developing flow okay inside a pipe, then we will be deriving
the energy equation and sub sequent boundary condition and we will giving a non dimensional
form to this one for uniform wall temperature case okay. Uniform wall temperature around a duct okay,
we will try to reduce the energy equation in to simplified form which can be solved
okay and that we will be doing for thermally developing region with constant wall temperature
boundary condition okay. And finally we will try to determine by this
equation energy equation will try to determine the Nusselt number for thermally developing
but hydro dynamically fully developed force convection around a duct okay having constant
temperature okay. So let me start with this schematic so here
we will be considering hydro dynamically developed but thermally developing flow, usually this
type of problems are called Graetz flow okay Graetz problem we can call okay. So here you see this is our pipe line let
say here we are having the center line okay, here we are having the center line and now
somehow the velocity boundary layer has developed and has taken actually a parabolic profile
okay. And then you see over here we are considering
that suddenly we are starting to heat over here the wall is getting a constant temperature
T=Tw okay, and here the fluid whatever was coming with the parabolic velocity profile
have a constant temperature Ti okay. And this is your axial detection of the duct
and this is you radial direction okay. The temperature profile will also become developed
after the certain length and what is the consequence in the further downstream sections those things,
we have discussed in the last two lectures. Here our major concentration will be over
here and the beginning of the heating section this region is called actually thermal entrance
region okay. You can see I have shown here the boundary
layer thermal boundary layer being developed by the way what is thermal boundary layer
this concept we have already discussed in lecture eleven okay. So this boundary layer will be developing
over here and whenever they will be merging at the center line beyond that we will be
having thermally developed region okay. So, let me try to first see what non-dimensional
parameter we can use as here you can find out this thermal entrance region is small
in length, we can take the scale of that one is r not which is nothing but he radius of
the tube okay. So we are non-dimensionalising z bar that
axial direction by r not which is nothing but the radius of the tube, so z equal to
z bar by r not okay, obviously the radius non dimensionalised radius can be taken as
r bar by r not for non-dimensionalization of temperature we are considering theta, so
theta is nothing but t minus tw by ti minus tw where ti is the inflow temperature and
tw is the wall temperature constant wall temperature okay. And as we have considered velocity boundary
layer is fully developed that means taking a parabolic velocity profile, we consider
that w which is non dimensionalised velocity axial velocity is nothing but w bar by w average,
w bar average is nothing but two into one minus r square a parabolic velocity profile
oaky. Now let me show you that what will be the
equations so the convection will be having only one term okay so axial convection term
because other two u and v those two things will be zero because it is thermally hydro
dynamically fully developed condition okay. So u and v is zero only w is having getting
a parabolic velocity profile like this so only singly term in our convection will be
remaining w delt delz okay and in the conduction side we will be having all terms but in this
we are only considering that radial and axial terms are remaining and we are having azimuthal
symmetry so theta directional terms will not be considering over here okay. So if you expand this equation so you will
be getting that if you expand then if you try to put all this non dimensional parameters
and replace the dimensional terms then you will be getting the convection is becoming
w average into w so this w bar is giving w average into w and this t is actually releasing
one ti minus tw and z is releasing rq and as a result delt delz is becoming del theta
delz okay. On the other hand in the right hand side this
theta is actually being replaced t is being replaced by theta and one ti minus tw is being
released and everywhere we are having second order term that means r square z square, and
here first order but multiplied with r so we can find out one r not square can come
outside because both z and r we have consider of the order r not okay, so ultimately this
equation we are getting for the hydrodynamically developed but thermally developing flow okay. So, if you simplify it little bit then you
can get that over here alpha and then r not into w average will be giving you peclet number
okay, so peclet number by 2 and here we can get the rest terms w into del theta del z
is equal to radial conduction terms and then the axial conduction terms okay. Now if you put the value of the w which is
nothing but two into one minus r square then we will be getting equation of this form okay
and subsequent boundary conditions definitely at r equals to one that means at the wall
we are having theta equals to zero because t becomes tw okay and at r equals to zero
obviously there will be no gradient of temperature so del theta del r equals to zero. Now at z tends to minus infinity that means
for before the pipe entry obviously theta will be one as at t becomes ti okay and that
z tends to infinity that means if you go for downstream in the pipe line so there we are
not knowing what is the value of theta because that will be depending on the length and so
we can write down that this theta is nothing but actually will be bounded okay. Next as our interest is lying in this thermal
entrance region so let us first do-little bit of scale analysis so if you see this thermal
entrance region, we will be actually having one minus r okay because this one minus r
is nothing but your thermal boundary layer thickness okay. So we can write down one minus r is the of
the order of the boundary layer delta okay, so this boundary layer delta is of the order
of one minus r. now if you use this one then you can write down one minus r square which
is a dominant term in the left hand side convection side in the equation. So you can get this becomes one minus r into
one plus r, one plus r is obviously of the order of one so you can get this one minus
r square is of order delta okay and definitely z we have already considered z bar is of the
order of delta so this z becomes the order of one okay. Now let us try to see that what are the orders
of all this term one by one so first if you see that del theta del r terms so del theta
del r so as theta is of order one it varies between zero and one so it can be of order
one r will be of order delta, so del theta del r term is of the order of one by delta. Similarly, del theta del z here theta is of
order one and z is of order one already we have mentioned so this becomes also of order
one okay, if you go for the second order derivative it will become one by delta square as del
theta del r is one by delta, and del square theta del z square obviously we will remain
same because z is of order one okay. So we have got all the terms okay order of
all the terms now if we try to see what is the convection order if you see the convection
order so del theta del z was actually of order one but one minus r square is of the order
of delta. So we find out multiplication of this one
is the of the order of delta okay. Let us see the radial conduction that means
these two terms if you see this two terms obviously we are finding out that this peclet
number can be taken in the radial conduction side in the conduction side rather, so it
is one by peclet number and then both the terms we are having one by delta square for
the first term and for the second term it is nothing but one by delta into r which is
of the order of delta. So, it is actually one by delta square term
okay, no actually this term is actually becoming one by peclet number order and this term is
becoming one/peclet number into one by delta square, so this term will be actually dominating
amongst these two terms. So, the magnitude or scale of the radial conduction
will become one by peclet into one by delta square okay. Then axial conduction if you see axial conduction
that already we have prove that this is of order one as one by peclet number came in
this side so it becomes of one by peclet numbers. So, here from you can get that between these
two conductions obviously this one is having higher magnitude so radial conduction dominates
over the axial conduction. If you equate this convection and conduction
side, we get peclet number is actually one by delta cube or we can write down delta is
of the order of one by Pe to the power one third this gives us some idea that what can
be our similarity variable. So, let us consider the similarity variable
so the similarity variable eeta we are writing one minus r into Pe to the power one third
okay. Now as we are considering one by r Pe to the
power one third then definitely this eeta will become of the order of one okay so this
is the beauty of this similarity analysis so by considering that what is the order of
delta or the boundary layer thickness we construct one variable call similarity variable which
becomes of the order of one okay. And already we have shown that z is already
of order one so we will get eeta and z coordinate now okay in place of z and r coordinate. Next let us see further that what will be
the value of r from this one as we have defined eeta in this fashion so little bit of site
change we can have r equals to one minus Peclet number to the power of minus one third into
eeta and subsequently one minus r square which is nothing but one plus r into one minus r
will be giving me Peclet number to the power of minus one third into eeta into two minus
that means one plus one minus Peclet number to the power of minus one third into eeta
and multiplication of this two terms will be giving me this one okay. Two Peclet number to the power of minus one
third eeta minus one minus of Peclet number to the power of minus two third into eeta
square okay. Let us proceed further for the derivative
of the theta terms okay so first del theta del r so if you do that theta is the function
of eeta, so del theta by del eeta into del eeta by del r so del eeta del r can be found
out easily by making derivative of this one, so this becomes nothing but minus Peclet number
to the power one third if you make the derivative of eeta with respect to R, so it becomes minus
Peclet number to the power of one third into del theta by del eeta okay, and second derivative
subsequently will be giving you Peclet number to the power two third del square theta del
eeta square okay, so once again you do the derivative with respect to this, once again
chain rule will be giving you Peclet number to the power of two third okay. Proceeding further if you put this derivatives
as well as the value of one minus r square in your equation that means in your equation
means over here rather than you will finding out this turns out to be like this. This is nothing but your two into one minus
r square okay, del theta del z two by Peclet number this is nothing but your del square
theta del r square and here you are having one by r so this is one by r term, one by
one minus Peclet number to the power of minus one third eeta and this one is nothing but
your del theta del r okay, and the axial conduction term remains like this two by Peclet number
into del square theta del z square little bit of simplification and site change will
be giving you like this four eeta minus Peclet number to the power one third eeta square
into del theta del z is equals to on the right hand side we are having two into delta square
theta del eeta square minus two Peclet number to the power of minus two third del theta
del eeta by one minus Peclet number to the power of minus one third eeta this is nothing
but coming due to r, plus two by Peclet number to the power of two third delta square theta
del z square okay. So you can see for large Peclet number what
we can do this term, this term, this term and this term can be cancelled because all
are carrying actually Peclet number to the power minus power, so you can find out only
remaining term is nothing but four eeta del theta del z is equal to two into del square
theta del eeta square that means it is nothing but two eeta del theta del z equal to del
square theta del eeta square okay. Let us see the boundary conditions also, so
we find out whenever eeta tends eeta is equal to zero now eeta is equal to zero means r
equal to one okay, so eeta equal to zero means r equal to one. Because we have considered eeta is equal to
one by r into Peclet number to the power one third okay, so we find out that at eeta is
equal to zero, theta is equal to zero because T is equal to Tw okay. Then similarly, as we have considered eeta
is equal to one minus r into Peclet number to the power one third though our boundary
condition was at r is equal to zero, del theta del r will be equals to zero so from there
we are getting that for large Peclet number, so Peclet number tends to infinity means eeta
tends to infinity because eeta is nothing but one minus r into Peclet number to the
power of one third so Peclet number becomes very big means eeta will be also very big. There we are finding out theta tends to one
okay, so and in case of the inlet we are having at z equal to zero, theta equal to one okay,
this is nothing but Ti minus Tw by Ti minus Tw okay. So we got the equation as well as the boundary
conditions two boundary conditions for eeta and one boundary conditions for z okay, so
this equation and sets of boundary conditions are actually called Leveque equation and the
solution of this one has been proposed by Leveque. So let us see how it can be solved at the
beginning what we will be doing, we will trying to find out the stretching variables for that
let us take eeta star equal to e to the power alpha one into eeta we do not know what is
the value of alpha one we need to find out. Similarly, Z star equal to e to the power
alpha two Z and theta star equal to e to the power alpha three into theta, so by using
this we will be trying to find out what are the values are alpha two, alpha one and alpha
three respectively, okay. So let us put all these values in this equation
and the boundary conditions. So first in the equation if you see there
was eeta, theta and Z so subsequently we are having e to the power alpha two minus alpha
one minus alpha three okay, and in the right hand side for the conduction side we are having
del square theta by del eeta square so from there we are getting e to the power two alpha
one minus alpha three and all the theta and eeta is turning out to be theta star and eeta
star, okay and if you see the corresponding boundary conditions B.C. boundary conditions,
so eeta tends to zero means obviously eeta star tends to zero. In that case you find out theta was equals
to zero now we are getting e to the power minus alpha three theta star equal to zero
okay, which gives nothing but for a finite value of alpha three this theta star equal
to zero okay, and for the axis eeta star tends to infinity means obviously eeta star tends
to infinity we are getting e to the power minus alpha three theta star tends to one
okay, and for the inlet boundary condition z star equal to zero we get e to the power
minus alpha three theta star equal to one, okay. Now from these two equations definitely we
can understand that we need to make alpha three equal to zero okay. Because it will be simplified version if alpha
three gets the value of zero then theta star becomes one and theta star becomes one over
here okay, in the inlet as well as in the axis okay. So let us now equate the co-efficient from
the equations okay, from the convections, conduction equation so we get alpha two minus
alpha one minus alpha three equal to two alpha one minus alpha three and substitute the value
of alpha three equal to zero so subsequently we get alpha two minus alpha one equal to
two alpha one okay. So from here we can get that alpha two equal
to three alpha one okay. So ultimately, we can then write down eeta
is nothing but e to the power alpha one into eeta, eeta star equal to e alpha one into
eeta, z star equal to e to the power three alpha one z and finally theta star equal to
theta as because alpha three equal to zero okay. So if you get so then it is very easy to find
out what can be my similarity variable so we can take eeta star by z star to the power
one third is equal to eeta by z to the power one third okay. So this correlation, this relation can be
found out from this two equations easily by eliminating e to the power alpha one okay,
so if you do so then we can get the similarity variable like this. Let us say similarity variable is zeeta. (Refer Slide Time: 22: 12) So we can write down zeeta is nothing but
eeta by z to the power one third and let us take a constant in front of that which is
capital A okay, so we need to evaluate the value of capital A and as we are having over
here you see theta star equal to theta so from here we can take theta is nothing but
a function of zeeta okay. So let us use this one and try to first get
what are the derivatives of zeeta with respect to eeta and zeeta, eeta and z, okay because
eeta is function of eeta and zeeta is function of eeta and z, okay. So here del zeeta / del eeta equal to A by
z to the power one third because here we are having A eeta by z to the power one third
okay, derivative with respect to z of zeeta becomes actually this form so A eeta, ok eeta
comes as constant and if we do the derivative of z to the power minus one third then we
get minus one third of z to the power minus four third so it will be ultimately giving
you minus zeeta by three z okay. So both the derivatives we have obtained of
zeeta then let us try to get the values of the derivatives of the non-dimensionalized
temperature theta. So first let us see the value of del theta
del eeta so del theta del eeta will be obviously as theta is a function of zeeta now, so del
theta del zeeta into del zeeta by del eeta okay, so here you see we are writing del theta
by del zeeta as theta dash and del zeeta by del eeta is nothing but A by z to the power
one third so we have got the value of del theta by del eeta okay, it was there in the
convection side, okay. Then double derivative if you see it was there
in radiations, in the conduction side sorry, it was there in conduction side. In same way if we do the double derivative
because in our conduction side single derivative term we have neglected for higher Peclet number
so we are only having del square double derivative of theta with respect to zeeta that means
del square theta del eeta square we are having, okay. So double derivative of this one okay, that
means once again if you the derivative with respect to eeta it gets theta double dash
A square by z to the power two by three okay. On the other hand, if you do the value of,
if you find out the value of del theta del z you will be getting it is nothing but theta
dash del zeeta del z, okay. So theta dash del zeeta del z is nothing but
minus zeeta by three z so that we can plug in over here, so we have got both del square
theta del eeta square and del theta del z so let us try to put that in the equation
okay, so in the governing equation if you remember earlier it was something like this
two eeta del theta del z equal to del square theta / del eeta square so here I have got
both the derivative values so let us try to put that over here quickly. So we can get two eeta then this is the value
of del theta del z and in the right hand side we are having actually del square theta del
eeta square, okay little bit of site change and modification it gives me A square theta
double dash equal to minus two third eeta by z to the power one third zeeta theta dash
here eeta by z to the power one third is nothing but zeeta by A okay, zeeta by A. So we can ultimately get that theta double
dash equal to minus two third zeeta square by A cube theta dash okay, and simplified
form of this one will be theta double dash plus two by three A cube zeeta square theta
dash equal to zero with the corresponding boundary conditions this boundary conditions
already we have seen theta equal to zero means at the wall it is actually equals to zero
and theta tends to infinity that means theta infinity equal to one it is at the axis, okay. Then let us see how this derivative, how this
equation can be integrated so for that we are writing one by theta dash and theta double
dash we are writing as del del zeeta of theta dash okay. On the right hand side we are having a constant
term not a constant term it is the function of zeeta minus two by three A cube zeeta square
okay. Now this can be integrated easily so this
I am writing as d/d zeeta of log theta dash and the right hand side we are having minus
two by three A cube zeeta square, okay. So if you integrate now one step then you
will getting log theta dash equal to minus two by nine zeeta cube by A cube plus a constant,
okay to evaluate the constant we will be using the boundary condition. So before that let us consider here you are
having two by nine A cube so this let us make a unified constant so for that we are considering
nothing but A equal to two by nine to the power one third so if you make A equal to
two by nine one third then A cube will becomes two by nine as the result you will be finding
out that this will become a constant, okay this will become a constant one. So we get if we choose A equal to two by nine
to the power one third so this co-efficient of zeeta cube becomes one okay, so ultimately
we will get that del theta del zeeta which is nothing but theta dash okay, is equal to
this constant we are considering actually logarithmic of B okay, so if we consider logarithmic
of B then we are getting ultimately theta dash equal to B into minus zeeta cube, here
A and subsequently two by nine to the power one third has actually given rise to one,
okay. Then once we get this one one step further
integration if we do from zero to zeeta we will be getting this is nothing but theta
(zeeta) minus theta zero put in the upper and lower limits and the right-hand side it
is nothing but zero to zeeta e to the power minus zeeta cube d zeeta with a B prefixed
over there, okay. So if we put the boundary condition that Theeta
zero is equal to zero at the wall so here you find out that theta (zeeta) equal to B
into zero to zeeta e to the power minus zeeta cube d zeeta, okay. So, as we have obtained the profile temperature
profile in terms of zeeta but only unknown is B. So let us try to plug in the other boundary
condition that means what happens at zeeta equal to infinity so at theta (infinity) it
is one which we have seen as boundary condition is actually equals to B into zero to infinity
zeeta will turn out to infinity now, okay e to the power minus zeeta cube d zeeta okay. So here I get what is the value of B which
is nothing but one by zero to infinity e to the power minus zeeta cube d zeeta, so
once I plug in this value over here theta becomes zero to zeeta e to the power minus
zeeta cube d zeeta divided by zero to infinity e to the power minus zeeta cube d zeeta okay,
so this we have obtained the profile for theta, preceding further if we have to evaluate this
integration let us consider zeeta cube equal to t okay. So once you see zeeta cube equal to t let
us take what is d zeeta, so d zeeta turns out to be one by three t to the power minus
two by three into dt okay. So if you put this value of d zeeta over here
and zeeta cube as t. Then we obtain this zero to infinity e to
the power minus zeeta cube d zeeta becomes zero to infinity e to the power minus t and
in place of d zeeta we can write down one third t to the power minus two third dt okay. Now here if you see this one third if we take
out, this is actually the expression of gamma function zero to infinity e to the power t
into minus two third dt. This is actually a gamma function for one
third. So we write down gamma one third, so this
is coming from mathematics once again gamma function is well knowing in mathematics okay,
so we get over here value of this integration of zero to infinity e to the power minus zeeta
cube del zeeta is actually one third gamma of one third. now using the rule of the gamma function this
one third gamma of one third can be written has gamma of four third okay, and the value
of gamma of four third if you see the gamma tables in mathematics you will be finding
out nothing but point eight nine three okay. So, we obtain theta zeeta which was earlier
was like this okay, now this value is nothing but gamma of four third or point eight nine
three. So, we can easily write down, it is nothing
but one by gamma of four third zero to zeeta e to the power minus zeeta cube d zeeta. So we have obtained the temperature profile
in this fashion okay. Next let us try to see the heat flux as the
subsequent heat transfer coefficient and nusselt number, so q the heat flux can be written
as k into del T del r at r equal to zero and if you convert r bar to r and t to theta,
it becomes k by r0 Ti minus Tw into del theta del r at r equal to one. So in this if you further put what is value
of heat transfer coefficient h which is nothing but q by Tw minus Tb okay, so this h if you
try to put then the other assumption we need to take as r is very small the volume integral
of Tb will be actually equals to Ti okay. So that means in thermal entrance length whose
length is very small, so if you do the bulk in that small length you will be finding out
very small amount of heat is been added. So actually, the bulk temperature will not
be changing much, bulk temperature can be considered very near to Ti. So here we are considering as r is very small
Tb will be nearly equivalent to Ti okay, so this comes from the very small thermal entrance
length consideration okay, so if you do so then here you see this Tb can be replaced
by Ti, so it is nothing but h equal to q by Tw minus Ti. But in some case where thermal entrance length
is finite that means a very big one, some fluid case it might happen there this assumption
not possible you have to find out the value of Tb in that case. But this assumption we are taking that Tb
is more or less near to this Ti for reducing our equations into a simplified form okay,
the assumption over here is that, very small r that means very small zone of thermal entrance
length okay. With this we can write down that if q by Tw
minus Ti is actually h so you can write down h is nothing but minus k by r0 into this derivative
del theta del r at r equal to one okay. So let us try to find out the nusselt number
further. So nusselt number we know which is nothing
but hD by k or D is the diameter h two r not by k, so ultimately Nusselt number becomes
minus 2 del theta del r at r equal to one okay. So del theta del r now if you try to convert
this r to eeta, which we have used for similarity variable then we will be getting two Peclet
number to the power one third into del theta del eeta and eeta equals to zero in this case
okay, because eeta was one minus r into Peclet number to the power one third okay. So if you do further then you see this value
del theta del eeta we need to find out which is nothing but del theta del zeeta into del
zeeta by del eeta okay. del theta del zeeta is theta dash and del
zeeta by del eeta is A by z to the power one third This already we have seen while deriving
the equations okay, subsequently del theta del eeta at eeta equal to zero, which is required
over here is becoming the value of A we have put as we have considered A is nothing but
two by nine to the power one third so we are putting over here two by nine to the power
one third theta dash zero by z to the power one third okay, so this value if you put over
here for finding out the nusselt number. The Nusselt number becomes two Peclet number
to the power one third multiplied by this whole term two by nine to the power one third
theta dash zero by z to the power one third. So as we have already seen that this can be
little bit modified and we know the value of theta dash zero. So what we can do little bit of change of
side over here giving me Nusselt number is equal to two into two by nine to the power
one third theta dash zero by z by Peclet number to the power one third, here we are having
Peclet number to the power one third here we are having z to the power one third this
is clubbing up over here as z by Peclet number to the power one third okay. and these factors
and along with this theta dash zero as we know value of theta zeeta equal to zero and
theta dash zero will be giving me actually one by gamma of four third. So gamma of four third value is known to me
so that if i have plug in over here the whole constant will become one point three five
seven so I get Nusselt number in the thermal entrance length is nothing but one point three
five seven divided by z by Peclet number to the power one third, in the thermal entrance
length region z will be function okay. So let us summarize in this lecture we have
understood governing equation for thermally developing but hydroynamically fully developed
forced convection over flat plate having constant temperature okay this is not correct convection
inside a duct having constant temperature and there we have found out this is the equation
two eeta del theta del z equal to del square theta by del eeta square for large peclet
number cases okay and the boundary conditions we have seen. Two boundary conditions for eeta this is at
the axis, this is at the wall and this is at the inlet z boundary condition required
because z is of order one okay over here. This solution we have called has Leveque solution,
and we have proceed further to find out what is nusselt number coming out to be for thermally
developing but hydroynamically fully developed forced convection inside a duct having constant
temperature. So this becomes nusselt number equal to one
point three five seven divided by z by Peclet number to the power one third okay. So this we have proved in the last slide okay. So just like your others lecture let us also
test how you have understood, so we are having 3 questions over here. first one goes like this in thermally developing
but hydroynamically fully developed region. Radial conduction is of the order we are having
four options one by Peclet number, one by Peclet number into one by delta square, one
by Peclet number into one by delta and Peclet number okay so obviously we have discussed
about this one and the correct answer is obviously b one by Peclet number into one by delta square. Second question is like this in thermally
developing region with constant wall temperature Nusselt number which statement is correct
statement a is constant, statement b depends on z, statement c depends on Peclet number,
statement d both b and c are true, so this already we have seen the expression of nusselt
number that Nusselt number depends on both z and peclet number so answer d is the correct
one. Last question is velocity profile in thermally
developing and hydroynamically fully developed region is four option we are having parabolic,
flat head, liner, or no conclusion can be made. We don’t know what is the velocity profile
so obviously you know as it is hydrodynamically fully developed the correct answer is parabolic
so this is also simple question okay, so with this we end this lecture. In our next lecture we will be discussing
the same thing thermal entrance region but we will be taking the conditions as constant
or uniform heat flux okay. So if you are having any query regarding this
lecture or any other general query about convection heat transfer please keep on posting on our
discussion forum thank you.

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