# Thermally and Hydrodynamically Developed Flow: Uniform Heat Flux

Hello, welcome in the 13th lecture of my course

convective heat transfer. In our last lecture we have discussed about slug flow okay where

velocity was almost constant okay, hydrodynamically the boundary layer is not developed. Here

we will be considering about thermal layer hydrodynamically developed flow that means

hydrodynamic boundary layer is already developed okay. That to in this lecture we will be considering

uniform heat flux, so that means the pipe is actually rapped up with some coil electric

coil let us say heater coil, let us say which is supplying uniform heat flux okay. So let

me first tell you that what we will be covering in this lecture. In this lecture we will be first deriving

conclusions for thermally fully developed flow in terms of the boundary conditions,

then we will be discussing or we will be reducing the energy equation into non dimensional form

for uniform heat flux case. So we will be considering only by uniform heat flux case

over here in this lecture. In the next lecture we will be discussing about uniform temperature

case. Solve the energy equation for obtaining temperature profile in terms of bulk temperature

and lastly determine the nusselt number for thermally and hydrodynamically fully developed

forced conviction over flat plate having constant heat flux okay. So, let me start with the topic as we are

only interested in thermal fully developed region, so as it is thermally fluid developed

so the heat transfer coefficient will not change with respect to time, with respect

to radius at least hz will be keeping as constant okay, hz is constant means as it is fuly developed

it will not change with respect to z, so hz is constant okay. As hz is constant let me

see that what are the other quantities, for example, bulk temperature okay. Bulk temperature is obviously the integration

over the radial plane, so here you can see this will definitely not depend on the radius

and so it is the function of z only okay. On the other hand wall temperature, wall temperature

obviously it will not depend on the radial plane okay, it will be only and it is only

varied at r not which is the tube diameter, tube radius okay. So T is actually function

of z, so here we are considering Tw is actually a function of z okay. So both we have got

as a function of z and hz is actually constant for thermally fully developed region okay.

Next let us see by definition this heat transfer coefficient h is nothing but Q by Tw-Tb okay,

wall temperature minus the bulk temperature okay. So here you see Q can be written as k del

T del r at r is equal to rnot that means at the wall okay. So here we get k del T del

r at r is equal to r not divided by Tw minus Tb, now here you see Tw and Tb both are function

of z, so what we can do this term we can take inside the del del r term, so there we have

done del del r of T by Tw minus Tb because this Tw minus Tb is actually not a function

of r okay. So as constant that can be taken inside. So actually, this T by Tw minus Tb

now it is becoming a function of r as T is function of r so this whole term is becoming

function of r okay. And as we have to make hz equals to constant

so obviously this has to be a function of r only, we will be putting the derivative

and then putting the limit r equals to rbar r equals to r0 and ultimately this hz will

become a constant okay. So let us define non dimensional number like this phi is equals

to T minus Tw by Tb minus Tw okay. So if you remember in our last lecture we have considered

Ti minus Tw here let us define with Tb minus Tw okay. Why because we want to make this

phi only as a function of r okay. So, this T by Tb minus Tw is actually a function

of r in the thermally fully developed region okay. Now if you define this phi in this fashion

then definitely phiW at wall the value of phi will become zero, because T becomes Tw

okay. Then let us see how the derivative looks like so T as for the definition of phi becomes

Tw plus phi into Tb minus Tw here I have showed the dependences based on the r and z coordinate

for different parts of this equation. Now if we go for the derivative with respect

to r first okay, if we go for the derivative with respect to r first this is the function

of z we need not to do, this will also be constant, this is not a function of r, here

it will be only becoming the derivative. So only you see this is remaining as constant

and this is becoming delphi delr okay. And now if we put r equals to rbar by r not then

this rbar needs to be transferred to r with 1 by r not multiplier with this term okay. So delT delr bar becomes this one okay. if

we proceed further for the heat transfer coefficient so for the heat transfer, from the heat transfer

coefficient we can see now this becomes as per the non dimensionalization of the temperature

it becomes actually –k by r not delphi by delr okay, so delT by delr okay you can get

from here delT by delr is actually delphi by delr multiplied by this term okay. So here

we are having already Tw minus Tb. Tw minus Tb and here Tw minus Tb will be cancelling

out with a minus sign okay. So this minus sign then K by r not delphi

by delr at one comma z this one is nothing but r equals to r not okay, r equals to r

not means rbar equals to r not means r equals to one okay. Then if you go farther for the

evaluation of nusselt number which is nothing but hD by k h into 2r not by k then from this

equation we get nusselt number is nothing but minus two of delphi delr at one comma

z okay. Now let us see for thermally and hydrodynamically developed flow with uniform heat flux which

we are supposed to discuss in this lecture, for that case we definitely as it is fully

developed U and V will be zero okay. So, my equation we will be turning into this

one in the left-hand side we are having the convection, but this time W is not a constant

and in the right-hand side we are having the conduction terms okay, azimuthal conduction

has been neglected. Now let us proceed further for constant heat

flux case, so if the heat flux is constant and as it is fully developed h is already

constant, then we can get obviously Tw minus Tb needs to be constant okay. So here q is

constant, h is constant so obviously it will be constant. Now if this is the case I can

show the temperature profile in this fashion let us say this is the z axial detection of

the tube and here I try to plot the T, so after this entrance length that is the thermal

developed region you can find out though due to supply of the heat flux temperature, wall

temperature is changing as well as the wall temperature will be changing keeping the gap

always constant. Tw minus Tb is always constant okay. So we will find out from here as it is constant

so d/dz of Tw will be obviously d/dz of Tb okay d/dz of T, d/dz of Tw is equals to d/dz

of Tb. Let us proceed further here let us do the derivative of this temperature in form

of phi with respect to z so delT by delz will remain here you see delT d/dz(Tw) because

this is the constant term now we have proved over here and phi r is not dependent of z.

So we get del delz of T is nothing but d/dz of Tw. And here we have shown that d/dz of Tw is

nothing but d/dz of Tb so ultimately d/dz of T becomes d/dz of Tb okay. Now let us try

to do little bit of balance of energy from the convection and whatever heat we have supplied

from there. So, let us say m dot is the mass attached with the pipe wall, so that is taking

the heat from the pipe, so m dot into Cp into dTb equal to q into 2phir into dz is the area

through which the heat is being supplied 2phir not is the perimeter multiplied by dz in which

the heating coil is wrapped up. So if you do this kind of balance then we

can find out dTb by dz is becoming 2pi rnot q by mdot into Cp here you see in right hand

side all terms are constant, so I can write down dTb by dz equals to constant. As dTb

by dz is equals to constant then definitely delT delz becomes constant, so delT delz is

constant over here okay, because we have already proved delT delz is nothing but dTb dz okay

so as this is constant definitely this term will remain constant. And if this is constant second derivative

of T over z will be obviously equals to zero, so that means we can obviously neglect an

axial conduction in case of thermally fully developed region. So in the equation last term we have dropped

this axial conduction term we have dropped and tried to non dimensionalize that one with

respect to w equals to wbar by wbar average okay. So as a result here wbar average came

okay. and we tried to convert the phi to theta sorry, we convert the T to phi and we have

written actually del delz of T to del delz of Tb okay which we have already proved. So

after little bit of site changing we can find out that whenever we convert this rbar to

r, rnot square also comes out, because we have considered r is nothing but rbar by rnot

okay. Now little bit simplification if you do then

del square phi del r square plus one by r delphi delr is becoming W average r square

by Alpha dTb dz into one by Tb minus Tw into w okay. So if we further do the simplifications

then we will be finding out that we are having over here two rnot by k into h into w okay

where this two rnot by k is nothing but your nusselt number okay. So this becomes nothing

but your nusselt number, here one minus sign is missing so this is becoming minus nusselt

number into w. So, my equation turns out to be this one del

square phi del r square plus one by r del phi del r equals to minus Nusselt number into

w okay. where this w is nothing but your parabolic profile hagen Poiseuille flow we are considering

so this is nothing but wbar is like this so w is becoming two into one minus r square

okay. So my final equation once I put the value

of w as two into one minus r square this is my equation, let us also see boundary conditions

so the boundary conditions are like this at r equals to zero definitely it needs to be

bounded so delphi delr equals to zero at r equals to one phi equals to zero that means

at the wall okay here r equals to one delphi delr equals to minus Nusselt number by two

this we have already proved few slides back, here we have proved, that at r equals to one

delphi delr is nothing but minus Nusselt number by two okay. So that will be also coming as

one boundary condition of my equations okay, so this is one equation at wall, because we

have already proved this one okay. Next let us try to see what is this value

of the phib already we have define phi as Tb minus Tw by Tb minus Tw so phib will be

obviously Tb minus Tw by Tb minus Tw which is nothing but equals to one, ok from the

concept of the phib, which we are discussed in eleventh lecture we can right down that

phib is nothing but zero to one W phirdr divided by zero to one W r d r okay. Now if this equals

to one then what we can write down as we know the profile of W which is nothing but two

into one minus r square this integration is very simple which gives me nothing but equals

to two okay, so nothing equals to half sorry, nothing but equals to half so this becomes

zero to one W phi rDr is equals to the half okay. Next let us try to solve this equation to

simplify it further let us consider that phi is equals to nothing but Nusselt number into

F okay, if you consider that my equation turns out to be like this cancelling Nusselt number

from both sides as we considered F equals to phi by Nusselt number so phi will be actually

Nusselt number into phi Nusselt number can be cancelled from both sides. So it becomes little bit simplistic in nature,

corresponding boundary conditions are like this is at the centre of the tube, axis of

the tube this is at the wall this is also at the wall coming from the Nusselt number

condition okay, this condition here also if you put phi equals to Nusselt number into

F then this Nusselt number will be cancelling from both sides ok so these are the boundary

conditions, let us now try to see the solution, this solution is now becoming simple. So, if you do it little bit of change of order

this becomes d/dr of r df/dr equals to minus two into R minus R cube okay, after once integration

ok, no not after once integration multiplication by R if you do the multiplication by R then

this comes out to be like this okay. Then let us go for the integration now so after

integration one step it will be involving one constant c one okay, which can be found

out from the boundary condition okay, at r equals to zero df/dr equals to zero so from

there we will find out c one is also equals to zero, if you do once step further the integration. Then we will be finding out F is becoming

one by two r to the power of four divided by four minus r square by two plus c two okay,

this can be also be found out putting this boundary condition at r equal to one F equals

to zero so c two gives three by eight ok. So finally, the F becomes one by eight into

three minus four r square plus r to the power four, let us now try to find out the Nusselt

number as per our consideration Nusselt number will be obviously Phi by f, because we have

considered phi equals to f into Nusselt number so this Nusselt number is Phi by f now this

can be written as zero to one r w phi dr divided by r w f dr okay. Now zero to one r w phi dr already we have

found out the value as half over here. Over here we have found out the value as half. We have found out the value as half over here

over here okay 0 to one w phi r dr is equals to half, so that we can put obviously in that

equation to get that Zero to one r w f dr is equals to what is the value in terms of

Nusselt number, one by nusselt number okay, now let us try to find out what is the value

of this integration as we know already the functional nature of f so what we can put

r f w is nothing but two into r into one minus r square multiplied by the f so this f we

have put over here little bit of algebra you do it will be becoming a polynomial of r. If u integrate from Zero to one then it will

be giving you eleven by nighty six okay. So further if we try to find out what is the

Nusselt number because we know that Nusselt number is nothing but half zero to one r w F dr,

so that zero to one r w F dr we have found out eleven by nighty six so this is the half

by eleven by nighty six and it comes out to be four point three six three six okay. So,

we have found out Nusselt number in this fashion so as we know the Nusselt number so the obviously

we can find out what is the value of phi, which is nothing but Nusselt number into F. So Nusselt number already we have found out

as forty eight by eleven so that we have put over here in front of the F, this was the

actually F which we actually found out so in this way we have actually calculated the

temperature profile okay, which is dependent on the radius as well as we have found out

the Nusselt number okay, so let us see what we have actually under stood in this lecture,

in this lecture we have actually discuss the governing equation for thermally and hydrodynamically

fully developed okay. So here this is very important fully developed

forced convection over flat plate having constant heat flux case, we have seen that the equation

comes out to be del square phi by del r square plus one by r del phi by del r is equal to

minus two Nusselt number into one minus r square. The boundary conditions we have established

as at r equal to zero del phi by del r equals to zero and at r equals to one phi equals

to zero okay. Also, we have seen that at wall it can be found out that del phi by del r

will be obviously minus Nusselt number by two. This we have proved from the equations

okay at the beginning, because we have understood that Nusselt number is nothing but minus two

del phi del r at r equal to r not comma z okay. And whenever we have tried to find out the

Nusselt number and non-dimensional radial distribution of temperature profile, we have

actually obtained Nusselt number comes out to be four point three six three six and the

temperature distribution becomes forty eight by eleven into three by eight minus r square

by two plus r to the power four by eight okay all these things we have found out for thermally

and hydrodynamically fully developed forced convection okay. And here probably this is mistake this is

not over the flat plate this is convection inside a tube having constant heat flux, Next

let me test your understanding what we have got in this lecture as usual we have three

questions, so first one in thermal fully developed region which quantity remains constant. As

it is thermally fully developed region we have four answers over here first one is H

and second one is actually Tw minus Tb, and third one you can say both and fourth one

we are saying none. Either you have to choose weather h or Tw minus Tb or both or none of

this h and Tw minus Tb, so obviously we have understood this we have elaborately discussed

in this lecture the correct answer is obviously h. So, h remains constant for thermally fully

developed region and when ever heat flux remains constant then some other futures also comes

into picture but only for thermally fully developed region only h will be remaining

constant okay. Next question is related to this one only

in thermally fully developed region with constant heat flux now we are implying extra condition

with constant heat flux which quantity remains constant okay, obviously as it is thermally

fully developed region h will remain constant and due to this constant wall heat flux we

have to find something more let us see the options the options are more are less the

same H Tw minus Tb both and none, so here obviously we can find out that we have proved

that in case of constant heat flux and thermal fully developed region Tw minus Tb is also

constant so the correct answer is both h as well as Tw minus Tb remains constant. The last question is in thermally fully developed

region with constant heat flux the value of the Nusselt number is three point six five

seven four point three six three six five point seven eight three one or none okay,

so this we have elaborately discussed and found out what is the Nusselt number so obviously

answer is four point three six three six okay. With this I am ending this lecture in our

next lecture will be discussing about thermally and hydrodynamically developed flow, but this

time we will be considering uniform wall temperature, in this lecture we have considered uniform

heat flux but here we will considering uniform wall temperature regarding this lecture if

you are having any query please do not forget to post it in our discussion forum thank you.