# 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.