Thermally and Hydrodynamically Developed Flow: Uniform Heat Flux

# 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.