# Thermal Entrance Region: Uniform Heat Flux

Dr. Arup Kumar Das

Department of Mechanical and Industrial Engineering Indian Institute of Technology Roorkee Hello welcome in 16th lecture of my course

convective heat transfer here in this lecture we will be discussing about thermal entrance

region once again but in this case we will be considering the pipe is having constant

heat flux boundary condition in out last lecture we have considered about constant wall temperature

condition here we will be considering the pipe is receiving constant heat flux so as

I have mentioned that we will be discussing about thermal entrance region with uniform

heat flux, so let me first show you that what outline will be following in my lecture so

first we will be deriving the energy equation. And boundary conditions okay in the form of

non dimensional numbers okay for uniform wall heat flux around a duct okay then we will

be reducing those energy equations into simplified form okay for thermally developing region

that we say entrance length along with the necessary boundary conditions which will be

evolving from the constant heat flux boundary condition okay then we will determine the

Nusselt number for thermally developing but hydro dynamically fully developed forced convection

inside a tube having constant applied heat flux will be finding out the Nusselt number

okay. So let us go inside this first I will be showing

schematically what situation we are going to study in this so as I have mentioned we

will be having a tube so this is a tube having center line this one r coordinate is radial

coordinate is in this side okay and this is axial coordinate in this side okay and here

we are considering hydraulically fully developed profile, so that means parabolic boundary

condition and there we are considering that in the tube we are giving constant heat flux. Let’s say the fluid whatever is coming that

is having some uniform temperature Ti okay, so with this as we are having hydro dynamically

fully developed flow so our w profile will be nothing but a parabolic profile and once

we non dimensionalise is w bar with respect to the average w so we will be getting w is

nothing but two into one minus r square okay where r is nothing but r bar by r not okay

next let us see for the wall heat flux which is a constant q we can write down q is nothing

but k del t del r bar at r bar is equals to 0 so which is coming from the wall okay. Boundary of the duct or tube okay so next

we will be using this boundary condition for getting the value for temperature non dimensionalization

this already we have shown in several cases so let us that same thing theta is equals

to T minus Ti Ti is the inflow temperature for the fluid which is constant okay and there

divided by q r not by k so q r not by k you can get from here so q r not will be multiplied

by k so it will be transforming into del theta del r okay. So if we take this type of non dimensionalization

for temperature then we can write down temperature in the form theta in this fashion okay so

T is equals to Ti plus q r not by k into theta let us find out the derivatives of temperature

with respect to r first because we will be requiring del T, del r bar over here so del

T del r bar becomes q now here q is constant okay and Ti is also constant so this term

goes to 0 so q r not by k that can be taken out so this q r not by k I have taken out

and then we are having del theta del r into del r by del r bar okay. So what we are finding out over here del r

by del r bar is actually 1 by r not so ultimately it becomes q by k del theta del r okay so

from there we get from this and this we get del theta del r at r equals to one is quals

to one okay. Next if we try to find out the other derivatives

and unlike this del T del r we will be finding out del T del z so if we do so it becomes

q by k del theta del z okay because we have considered T is equals to Ti plus q by k r

not into theta so we will be finding out that it is becoming and we have considered z is

is equals to z bar by r not okay because it is thermally developing zone so entrance length

so the axial distance will be of the similar order of the radius okay. So we have seen that if we do the derivative

with respect to z bar so it will become q by k into del theta del z okay so this is

del theta del z and then once again we will be having del z by del z bar so del z bar

gives one r not and r not and here q not that r0 will be cancelling out and subsequently

we will be getting q by k del theta del z okay so if we do second time derivative we

will be getting similarly q by k r not x del square theta del z square okay and if we do

second time derivative of T with respect to r bar so then we are getting q by k r not

del square theta del r square. So all this benefits of the temperature let

us but in the energy equation so energy equation is like this we have seen as it is thermally

fully developed so u component and v component of the convection is not coming only the axial

component w component is coming into picture and in case of conduction we are having the

radial conduction and axial conduction azimuthally conduction we have not considered over here

okay azimuthally symmetry has been assumed okay then let us see if we put all this non

dimensionalization. That means convert T to theta z bar to z and

r bar to r then we get over here once again w can be converted to w bar can be converted

to w by multiplying w bar average then we can find out that the equation actually simplifies

to w average w bar average into w, q by k into del theta del z in the left hand side

for convection in the right hand for conduction alpha by k into q by r not okay and then we

are having the radial conduction non dimensionalized radial conduction plus axial conduction term

okay further simplification cancelation of terms form both side we can get w average

r not by alpha which is nothing but k by del cp okay k by row cp alpha is over here once

again. So we can get over here some non dimensional

number which is nothing but actually Peclet Number. Peclet Number if we define based on

the diameter so Peclet Number by two will be coming over here okay and as w is actually

a parabolic velocity profile so that can written as square into one minus r square okay. So ultimately we get very simplified equation

like this one minus r square del theta del z is equals to one by Peclet Number plus radial

conduction plus axial conduction okay. Let us also see the boundary condition so

obliviously at the wall r is equals to no at the axis r is equals to 0 our del theta

del r will be 0 that means there will be no gradient of temperature across the axis in

the radial direction okay so this gives us the symmetry boundary condition sort of and

then r is equals to one is nothing but your wall boundary condition so in the wall boundary

condition we are having actually del theta del r is equals to one which is nothing but

constant heat flux boundary condition okay we have chosen theta such that this boundary

condition actually reduces to a simplified form del theta del r is equals toone okay. And then for the axial directions you are

having z tends to minus infinity and z tends to infinity that means z tends minus infinity

means it is far before the entry of the pipeline and z tends to infinity means in the downstream

or the pipe line so in case of z tends to minus infinity we will having theta to 0 because

it becomes Ti minus Ti divided by Ti minus Tw so it becomes 0 and in case of z tends

to infinity through it is supplying heat via the heating coils as a function of you know

constant heat flux but temperature definitely we will not be going to infinite it will be

always the finite so let us have theta is finite boundary condition at z tends to infinity

okay. So with this equation and boundary conditions

two boundary conditions of r and two boundary conditions of z we can describe the thermally

developing region inside a pipe line okay which in which we have considered that hydraulically

full developed fluid is going on okay, next let us try to see some scale analysis so for

the thermally developing zone we need to see that what is first the scale of your axial

direction and radial direction. As it is thermally developing zone so the

zone will be very small obviously that will be of the order of r not okay so z will be

coming of the order of one because z is nothing but z bar by r not okay and in case of radial

direction one minus r is nothing but r not minus of r bar divided by r bar so that will

be coming in the form the boundary layer thickness delta not okay so this two scale once we decided

then we can find out quickly the other scales for example from the boundary conditions here

you see del theta by del r needs to be of the order of one. So as we have decided r is of the order delta

not so obviously theta needs to be of the order of delta not okay because this del theta

by del r needs to be order one okay so if r is of the order of delta not obviously needs

to be order if delta not okay so we have already found out z r and theta order so let us find

out the rest terms of the equation so first let us start with one minus r square so one

minus r square will be one minus r into one plus r okay so here we can get it will be

also of the order of delta . Why because one plus r is actually of the

order of one okay so one plus one it would be order of two so basically this one minus

r square will be of the order of delta not here this one minus r is actually determining

what is the order okay next let us see that what is the order of convection. So first in this equation we have the convection

so convection was w x del theta del z okay as we have seen that part of convection this

w is nothing but two into one minus r square so one minus r square is order of delta not

so this w can be written in the order of delta not and del theta by del z, z is of order

one and theta is of order delta not just now we have said so we get the convection is of

the order of delta not square okays then radial conduction if you think about so in this equation

if you see the radial conduction is having one by Peclet then theta order and r order

square okay so here also r order square theta is order over here okay. So we will be finding out radial conduction

becomes one by Peclet Number into delta by delta not square okay so radial conduction

order we have found out let us see the axial conduction in case of axial conduction it

is nothing but one by Peclet number then let us go back to the equation. one by Peclet Number then order of theta and

order of z, order of z is one that’s whole square will be also of order of one and theta

will be of order delta not, so ultimately, we get one by Peclet number. one by Peclet

Number into delta not by one okay now you see this order is actually smaller compared

to this one, okay. Because delta is very small okay so we can get the conduction order is

actually this one which is the radial conduction, axial conduction can be neglected in compare

to the radial conduction for small delta obviously, okay. So what we can take let us equate the

convection order and radial conduction order. So, delta not square which was the conviction

order and del one by Peclet Number delta not by delta not square which was the radial conduction

order, if you equate from here we can get delta not is of the order if Peclet number

to the power minus one third okay so we have got the boundary layer thickness order, okay.

So, it will be helping us to construct the similarity variable. So let us try to have similarity variable

in case of our thermal entrance terms region okay, so here once again schematically I have

shown the thermal entrance region so up to this we will be having the thermal entrance

region, so in that as we have already taken z is of the order of one and earlier we have

taken one minus r is of the order of delta not and in the last slide by equating the

convection and conduction order we have shown that delta not is actually Peclet number to

the power minus one third so one minus r can be written as Peclet Number to the power minus

one third, okay. And let us take the similarity variable eta

in this fashion, eta is nothing but Peclet Number to the power minus one third into one

minus r okay, so r becomes one minus Peclet Number minus one third into eta so eta is

our similarity variable, so if we take so here I have clearly showed what is the variable

eta so if we take so then quickly we can try to find out the derivatives of theta which

will be useful for finding out the derivatives of temperature, okay theta. So let us find

out first del eta del r. So del eta by del r simply it becomes minus

Peclet Number one third okay and then subsequently if we try to find out del theta by del r,

del theta by del r will be nothing but del theta del eta into del eta by del r, del eta

by del r just now we have found out has minus Peclet Number one third so it becomes del

theta by del r becomes minus Peclet Number to the power one third into del theta by del

eta , okay. Then let us do one more derivative if you do once more derivative del square

theta by del r square it becomes once again Peclet Number to the power one third into

del square theta by del eta square we need to plug in this del eta by del r once more

over here from this del theta del r to del square theta by del r square, okay. So both

the radial conduction parts we have obtained. Next let us find out the value of one minus

r square which is there in the convection side due to the parabolic velocity profiles,

so it becomes actually one minus r into one plus r which is nothing but Peclet Number

to the power one third into eta multiplied by minus Peclet Number to the power minus

one third into eta , this two is nothing but one plus one minus Peclet Number to the power

minus one third into eta which is one plus r and this is actually your one minus r okay.

So, after that if we try to put everything in this equation then we simply get Peclet

Number to the power minus one third eta, okay. So into two minus Peclet Number to the power

minus one third eta this is nothing but your one minus r square we have derived over here,

into del theta del z is equals to one by Peclet Number and then in the right hand side we

have already evaluated del square theta by del r square and del theta by del r let us

put those so del square theta by del r square is nothing but Peclet Number to the power

two third into del square theta by del eta square and r we have written as here r we

have written as one by one minus Peclet Number to the power minus one third into eta okay.

And del theta by del r is nothing but Peclet Number two the power one third into del theta

by del eta, okay. And last term remains as it is over here because

Z is of order one okay so if you simplify this equation little bit then we get this

type of equation and we find out that in three terms here in axial conduction in this del

theta by del eta term and a part of this del theta by del z term we are having Peclet Number

minus power okay. Now for large Peclet Number limit okay, so taking Peclet Number very large

what we can do, those terms we can cancel out and make our equation simplified, so in

case of large Peclet Number we can drop down this terms and we can write down two eta del

theta by del z in the left hand side coming from the convection and in the right hand

side del square theta by del eta square coming from the radial conduction only.

So this becomes very simplified equation for large Peclet Number cases, okay. And let us

see also the corresponding boundary conditions, so here first boundary condition we had earlier

in terms of r it was earlier at r is equals to one which is at the wall del theta by del

r is equals to one which was actually constant heat flux boundary condition, so from their

first we need to convert to your eta boundary condition. Because we have already seen that eta is equals

to one minus r into Peclet Number one third so here if we put the value of r is equals

to one then we get eta is equals to 0 okay, so at eta is equals to 0. The boundary condition comes out to be the

boundary condition comes out to be though theta will not be changing but due to this

r to eta one Peclet Number one third will be coming out so the boundary condition will

transform into Peclet Number to the power one third into del theta by del eta is equals

to one, okay. So, this becomes the equation and this is the boundary condition, now you

see the boundary condition is Peclet Number to the power one third into del theta by del

eta is equals to one is complicated one. Let us try to replace theta by some simplified

one so that this boundary condition can be written in some better format, okay. So let

us take theta is equals to Peclet Number minus one third into theta one, this will not change

the equation because in equation both the sides we are having theta so this Peclet Number

to the power minus one third can be cancelled from both the sides, but it will be definitely

helping in this boundary condition because in boundary condition we are having Peclet

Number to the power one third into del theta by del eta is equals to one. So, once we put this theta one is equals to

theta by Peclet Number minus one third those Peclet Number to the power one third and Peclet

Number to the power minus one third from here we will be cancelling out and it will become

simplified, okay. So, let me show you after putting this theta is equals to Peclet Number

to the power minus one third with theta one what form of boundary condition we get. So,

this is the equation, now you see. (Refer Slide Time: square0:40) theta has been converted to theta one and

Peclet Number to the power minus one third has been cancelled from both sides okay but

boundary condition it is very surprisingly changing to at eta is equals to 0 which is

nothing but r is equals to one, it is changing to very simplified form del theta one by del

eta is equals to minus one, earlier it was Peclet Number to the power one third into

del theta del eta is equals to minus one okay, so this simplified equation and boundary condition

we get, apart from that we are having some other boundary conditions also. Let us see, so first that r tends to del,

theta tends to 0 or so this was a boundary condition earlier so from here we get if eta

has eta tends to 0, theta one tends to 0 okay, and at the inlet as z tends to 0, theta tends

to 0 okay, so this is being converted to at z tends to 0, theta one tends to 0. So now

we have got theta one’s equation and one, two and three boundary conditions for theta

one, two for eta direction and one for z direction, okay. Let us now try to get the similarity

variable once more in between minus between eta and z. So first let us try start quantities eta star

is equals to e to the power alpha one into eta , z star is equals to e to the power alpha

two into z and theta one star is equals to e to the power alpha three theta one, okay

and first we need to find out all this derivatives we have to replace all this theta one to theta’s

in equation as well as boundary conditions for that let us put this start quantities

in the equation first so here you can get in the left hand side as we are having eta

theta one and z we are getting e to the power alpha two minus alpha one minus alpha three

in the convection side in the conduction side we are having delta square it is a mistake

over here it will be delta square, delta square theta one by del eta square so it becomes

actually del square theta one star by del eta square and it releases e to the power

two alpha one minus alpha three okay. So, then the boundary condition if you see the

very important boundary condition it becomes eta star is equals to 0 it will be e to the

power alpha one minus alpha three into del theta one star by del z star is equals to

minus one, okay. Now we are having this is the equation and

this is the boundary condition from here easily we can tell that okay, if we take alpha one

is equals to alpha three then this boundary condition becomes very simplified, okay. And

other boundary conditions at eta tends to infinity and z is equals to 0 remains as usual

similar theta one star tends to 0 and theta one star is equals to 0 okay, so one relationship

between the constants alpha one, alpha two and alpha three already we have obtained from

the boundary condition alpha one is equals to alpha three. Now let us see the equation coefficients,

if you equate the power of the e to the power, if you equate the coefficients of this left

hand side and right hand side terms and get a relationship between alpha’s then you

can write down from there. Alpha two minus alpha one minus alpha three

which was the power of e in the conviction side is equals to two alpha one minus alpha

three which is the power of e in the conduction side, okay. So from here we can get if you

further simplify alpha two is equals to two alpha one minus alpha is equals to alpha one,

here alpha three has been cancelled from both the sides okay and if you further proceed

you will get alpha two is equals to three alpha one, so we have got two relationships

one is alpha two is equals to three alpha one and another one is alpha one is equals

to alpha three, using this let us try to eliminate alpha two and alpha three and write down the

star equations in terms of alpha one only, so we get eta star is equals to e to the power

alpha one eta z star is equals to e to the power three alpha one z because alpha two

is actually three alpha one and theta one star is equals to e three alpha one theta

because alpha three is equals to alpha one okay. Now if we put that from here if we try

to get the similarity parameter zeta so these two equations will be helping me so let us

see from these two equations, we can get eta star by z star to the power one third is actually

equals to eta by z to the power one third okay. And from this two we can easily get that theta

one star by z star to the power one third is equals to theta one by z to the power one

third okay and this can be also written as we are seeing that the order is more or less

same it can be also written as this will be a function of eta by z to the power one third

so ultimately let us have the similarity variable zeta actually is equals to eta by z to the

power one third multiplied by a constant A which needs to be determined, okay. So we

have actually written the value similarity variable side by side let us also take theta

one is nothing but B into Z the power one third f of zeta okay. Its comes from here

okay so B into one third fof zeta where zeta is this one okay. So, after defining this

zeta and theta one in form of in form of zeta let us first try to see what is the derivative

of zeta with respect to eta and z okay. these are very simple make the derivative of this

one with respect to eta first considering z constants and this will be by considering

a eta constant okay. Once you do these two derivatives you will

be getting. Using this to let us find out the values of temperature derivatives. So

first {\displaystyle \partial }

{\displaystyle \partial }{\displaystyle \partial }del theta one by del eta so if you do a chain

rule then you will be finding out the standing out to be it is standing out these AB f dash

okay so AB f dash to the turning out to be. Here we have using these del zeta by del eta

which we have to found to over here and del theta one by del eta will be making derivative

of the one should be B will be coming and A will be coming from here as the result it

is AB f dash okay. if you proceed further second derivate of theta with respect to eta

will become A square B is z it the power one third f double dash okay. And if you find out what is the derivative

of theta one with respect to z from its dependence on z over here then will be getting the requires

a it requires derivation of two multipliers so that if you do that one by one by keeping

one constant and the being the other derivative and then typing the second one constant who

doing the derivative of the first one who is get the simplified form like this. This

is nothing but B by three into z to the power two third into f minus zeta f dash. So we

have obtained both the temperature derivatives of theta one okay? Temperature derivatives

of theta one a with respect to and z respectively. Second derivative and first derivative of

z. these there energy equation. So, let us try to put all these values in the energy

equation. So, have I have told two eta and this is nothing but your del theta one del

z okay. So, del theta one del z okay? So, these are all put over here and I mean right

hand side we are having del square theta one then del eta one let as put it over here okay.

Now here you see in this equation we can see that B can be cancelled. So ultimately you

get equation having A square only okay? And a and you see little bit of a modification

of this eta by z to be power one third which is the zeta we can get a simplified form like

this okay? If you if you change the sides of this one then it will be getting f double

dash plus two zeta by three A cube into zeta f dash minus f is equal to 0 okay. Now you see a boundary conditions of sequentially

changing to at zeta is equal to zero AB f dash to is nothing but del theta one by del

eta. So we have actually seen that here AB f dash is nothing but del theta one by del

theta one by del eta which is actually equal to which is actually equal to minus one so

AB f dash is becoming minus one okay. And a obviously the other boundary condition that

as zeta tends to minus and zeta tends to infinity f is becoming 0 okay. So this the equation

and we are having a two boundaries conditions over here okay. If you proceed further and try to find out

what is the value of A to make this equation simplified in loop we can choose the value

of a in this fashion three A cube is equals to one if you choose when you see three A

cube is equal to one if you choose then the equation will become very simple okay? So,

you can work down this three A cube. So, in that case in that case sorry apart from this

one we can also take this AB equals to one then the boundary condition will be also very

simple. At zeta equals to zero f dash is equal to

minus one okay? So, we have got three A cube equals to one and AB equals to one. If you

choose like this then from here we can get individual values of A and B like this. A

is equals to two third to the power one third and B is three by two to the power one third

okay. Now let us plug in all these things in our equation so our similarity variables

become eta by three by two three z by two to the power one third and theta one becomes

three z by two to the power one third f okay. So earlier here we had A and B respectively.

Once you get the value of A and B we can get the actual value of zeta and theta one over

here okay? And in case of equation the equation becomes now very simplified in loop have we

put three A cube is equals to one and the boundary condition also looks very simple

as we have put AB equals to one so this my equation and will having two boundary conditions

in this fashion okay. now for finding out f this is not a very simple equation as we

are second order first order and zeroth order. So it is a, we need some numerical simulation

okay? You have to find out if numerical simulation. And if you try to get back the value of theta

as we have taken the as we have taken the equation over here from theta one we can go

back to theta as we have taken theta is nothing but a Peclet number to the power minus one

third into theta one. So, it is the go back here so theta becomes three z by two Peclet

Number to the power one third into f. so once you get the value f by the numerical simulation

of that then you can get the temperature distribution one dimensional temperature distribution theta

in this fashion okay. Next let us try to get the value of Nusselt

number so in order to do so first will be coming the heat transfer coefficient h. which

is nothing but q by Tw minus Tb okay? if you proceeds further so you will be getting h

r not by k which is which can be you know the Nusselt number to into h r not by k is

Nusselt number so there you can find out this is nothing but q r not by k by Tw minus Tb

and Tw minus Tb can be reduce Tw minus Ti minus of Tb minus Ti so now this can be written

as a theta w and this can be written as theta b because Tw minus Ti by q r w by k is nothing

but theta w and Tb minus Ti by q r not by k is theta b okay. So we get h r not by k is nothing but one

by theta w minus theta b okay? Proceeding further Nusselt number will be nothing but

two into this factor so this becomes two by theta w minus theta b okay. Now a theta b

about theta b this will be nothing but w theta d r integration from zero to r and then w

d r from 0 to r. as it is thermally fully developed so w will become actually two into

one minus r square so two two can be cancelled from denominator and numerator okay. And a we see we see that theta w okay the

theta w as we have shown in the previous one theta variable so theta was three z by two

Peclet Number to the power one third and then f of zeta okay. now as we are at wall zeta

obviously zeta will be zero. So, we get theta w is this fashion. So here you see we have

theta w here we have got theta b which is 0 so we can easily find out the Nusselt number.

So Nusselt number becomes theta w is nothing but this one so this quantity we have return

minus theta b which is zero so this is actually returns two by theta w okay. Now once you get what is the value of this

a a f not okay so after simulation of the equation and the boundary condition, we have

shown governing equations so then will be getting the values comes out to be one point

six three nine okay? So, if the Nusselt number becomes one point six three nine by z by Pe

to the power one third. So here also we can see Nusselt number function of z and the Peclet

Number over here okay. So we this I end this lecture let us summarize

what we have learnt so first we have seen the governing equation for thermally developing

but hydrodynamically fully developed forced convection inside a tube having constant applied

heat flux. So, this is very important constant heat flux so the equation we can reduce to

this from two eta del theta del z is equal to del square theta del eta square okay. Corresponding

boundary conditions so this is the boundary condition at wall okay. del theta del r is equal to one and these

are the one boundary conditions at the axis okay. At this is the boundary condition at

the entry. Once you reduce to subsequent stages via similarity variables and theta two theta

one finally we get f double dash plus zeta into zeta f dash minus f is equal to 0 with

the boundary condition f dash 0 is equal to minus one and f infinity is equal to 0. We

have also proceeded further to show that relationship of Nusselt number with respect to z. So Nusselt number for thermally developing

but hydrodynamically fully developed forced convection inside a tube having constant applied

heat flux comes out to be Nusselt number is equal to one dot six three nine z by Peclet

Number to the power one third okay which is dependent on z okay. So, this you have learnt

in this lecture so I will after this one let us see that how for you have understood in

this lecture so let me test to the understanding where three questions are usual. So first one is in thermally developing but

hydro dynamically fully developed fully region inside a duct having constant heat flux, radial

conduction is of the order you have to give me the oreder of radial conduction four options

are there I think already we have understood which one is the correct answer. So, options

let me tell you once again one by Peclet Number one by Peclet Number into one by delta square

here we are having one by Peclet Number into one by delta and finally Peclet Number. So

obviously you have got the correct answer that is one by Peclet Number is the correct

order okay. So this we have discussed also in this lecture.

Next one in thermally developing region Nusselt number depends on both Peclet Number and z

for which case. Though in this lecture we have discussed about constant heat flux but

let me ask you that your Nusselt Number depends on both Peclet number and z for which case.

Constant wall temperature constant heat flux none of these case a will be the to giving

you the dependence on the Nusselt number and Peclet Number and z and fourth option is both

a and b are true. If you go through this lecture and we see the previous lecture also then

probably you have understood correct one is both a and b are true. Actually, in thermal in terms region the Nusselt

number will depend on both Peclet Number and z. so that is why it will not be depending

on the whether the wall constant temperature or constant heat flux that will not be mattering.

It will be mattering whether it is thermal entrance region or not okay. So, it will be

having z dependence as well as in the Peclet Number dependence. Last question is like this

in thermal developing but hydro dynamically fully depend region inside a duct having constant

heat flux one minus r is of the order this is very simple one minus r obviously we have

used for our similarity variable. So one minus r is the order of four options

here we having Peclet Number to the power minus one third Peclet Number to the power

one third Peclet Number and one by Peclet Number. Very simple probably you have understood

correct answer. So correct answer is the first one Peclet Number to the power minus one third

okay so with this end this lecture a in the next lecture will be discussing a special

case which called a Rayleigh Benard convection which we can observe in case of flow inside

a duct inside two parallel plates actually okay. The dimensional parallel plates there will

be understanding what type of convection sails are there are end going for the stability

analysis of that one okay. if you have any query regarding this lecture and any other

general query about convective transfer please keep on posting discussion for on thank you.