Forced Convection over a Flat Plate: Uniform Heat Flux

Forced Convection over a Flat Plate: Uniform Heat Flux


Hello welcome in the 7th lecture of the course
convective heat transfer. In this lecture we will be discussing about forced convection
over a flat plate. But this time we will be considering uniform heat flux is applied at
the plate okay. In the previous lectures we have discussed the wall was at constant temperature,
here the wall will be supplying uniform heat flux okay. So let me show you what things
we will be discussing over here. First we will be studying the behavior of
thermal boundary layer over a flat plate which supplies constant heat flux to the adjacent
fluid. Then we will be discussing or we will be determining the parameters for a non dimensionalization
of temperature, because in earlier case temperature was straight away temperature non dimensional
came straight away because the wall temperature was known. But in this case the wall temperature is not
known, because we are supplying the constant heat flux, so what will be the parameter for
non dimensionalization that we need to see over here okay. Then we will deriving the
non dimensional energy equation along with the boundary conditions for uniform heat flux
case okay. We will be showing both the limits high and low prandtl number in this case okay.
And we will be mentioning the energy equation if we consider the viscous dissipation with
constant heat flux. Now let me show you the situation what is
happening over here so this is the flat plate let us say okay. Free stream velocity is U
infinity over here and the free stream temperature is T infinity but in place of wall temperature
Tw now we are having the heat flux known which is qw and it is constant in magnitude okay.
So let us consider that we have to non dimensionalize the temperature, so earlier we have done T
minus T infinity by some temperature drop okay, earlier it was Tw minus T infinity,
but in the absence of Tw my concern is what would be this delta Te okay. So let us try to find out that one, for that
we will be starting from the boundary conditions, because now the boundary conditions we will
be varying earlier it was T at wall was Tw, now it will be minus k del T del Y at Y=0
that means the wall here Y is in this direction positive direction minus k del T del Y is
actually equals to a constant we are taking that one as Qw okay, so this is the boundary
condition at wall we are having right. So if you do the scale analysis of this one
so for T we have seen it is a border of delta Te which is for still unknown, so delta Te
we have taken, for Y we have taken L okay so delta Te by L is actually Qw okay. So this
is the order of the equation okay, if you non dimensionalized Theta and Y in this fashion.
So Ybar has actually replaced over here by Y, because Y is actually we have considered
Ybar L by L in the previous lecture okay. So from here you see this became very complicated
boundary condition, we can reduce this boundary condition in a simplified form just if we
can shade of this part and we can simply write down del theta by del Y is equals to some
constant that will be convenient for us. So we can choose this deltaT in such a fashion
that it captures all this other constants and make this boundary conditions very simplified. So in order to do so what we are writing del
theta by del Y at Y equals to zero is equals to minus one we want let us say okay. So we
want to shade up all these coefficients. So in that case delta Te needs to become qw L
by K okay. So here we have got some idea of the temperature scaling okay in terms of the
wall heat flux qw okay. So this temperature non dimensionalization scale we will be using
for the rest part of this lecture okay. So have i had showed so theta will be T minus
T infinity and delta Te we have replaced by qw L by K over here okay. So let us try to see that how this is affecting
the equations, energy equations. So we will be starting from the generalized equations
so this is the dimensional form of the equation having conduction in the right hand side,
and convection in the left hand side remember here we have not considered the viscous dissipation
part. Now if you try to do the non dimensionalization in that case it is becoming one by Re Pr. So what we have done essentially over here
is that this X we have converted X bar we have converted to X, Y bar we have converted
to Y as a result this Re Pr came over here okay. So Alpha is actually giving rise this
Peclet number or Re Pr okay. Now let us try to go for this steps variables, so Y is becoming
small y and y is becoming small y into Re to the power half and capital V is becoming
small v into Re to the power half already we have seen in the previous lecture. So if we do so then we can actually neglect
the stream wise conduction term okay. So this one by Re Pr del square theta Theta del x
square can be cancelled and finally this equation already we have shown in the previous lecture
okay where stream wise conduction has been neglected okay. Important thing will be boundary
condition earlier we have seen that del theta by del y was minus one, now as we have used
capital Y and small y relationship like this so here Re to the power half is coming in
front of that. So Re to the power half del theta by del y is equals to minus one, this
is capital Y remember okay. And far away from the plate the condition will not be changing,
so theta infinity is actually zero and ending towards zero. Now as we have done in the previous lecture
also that eta will be taking as capital Y by root over of two x similarity variable
and using this what we can do, we can write down U equals to f dash of eta and V is actually
one by by root over of two x f dash eta minus f okay, so this we can take from usually mechanics
knowledge okay. \
Then here let us define theta as a function of eta in this fashion theta is nothing but
your temperature to find non dimensional temperature profile in X and Y coordinate and the we are
writing root over of two x by root over of Re into phi of eta. So all this X and y is
coming in eta okay, though root x is over here in this equation okay. Now if you try
to do the differentiation of this one with respect to small x and small y because those
times are there in your energy equation as we have shown in the previous slide. So here you see del theta by del y first it
becomes det theta by del eta into del eta by del capital YTheta which is nothing but
one by root over of two x okay. So this det theta by del eta actually becomes root over
of two x by root over of Re into phi dash of eta because these are not function of eta
okay so only this is the function of eta so this became phi dash okay. And one by root
over of two x remains over here okay, so and these two terms can be cancelled from here
in this equation it simplifies to phi dash by root over of Re okay. So it can be written
phi dash is equals to Re to the power half del theta by del y okay. So boundary condition
whatever we have showed in the last slide so the boundary conditions as we have seen
here it is del theta by del y at Y is equals to zero is minus one and which we have shown
over here after making small y to capital Y Re to the power half into del theta by del
y is equals to minus one so that can be now converted to theta can be converted into phi
using this equation. So Re to the power half del theta by del y
is equals to phi dash so phi dash will be now becoming minus one okay. So phi dash to
zero is actually minus one is the boundary condition at the wall okay. And away from
the wall remains same so phi tends to zero at whenever eta tends to infinity okay. Also
we can go for the higher order derivatives of this term which will be requiring for the
energy equation, so del theta del y was here in this fashion if you do the derivative once
more with respect to capital Y so this become one by root over Rex into phi double dash
okay. And if you go for the del theta by del x this
is also a term in the convection side of the energy equation, so this derivation gives
you the final form of this fashion. So here you can find out the final form is coming
as root over of one by two Rex into phi minus eta phi dash okay. So these are simple states
you have to do the derivation of theta with respect to X we are having two terms actually
root two x and phi both are function of X. So if you see over here first we have kept
phi I have done the differentiation of root two x and here root two x has been kept constant
and phi derivation we have done with respect to first eta and then eta is derivation with
respect to X okay. And further simplification has given rise
to this one okay. So all three terms we have found out del theta del x, del theta del y
capital Y and del theta del square theta del capital y square so let us put in the equation
okay. So energy equation which we have already shown. So first this is your U, U is nothing
but f dash okay and this term is actually your del theta by del x okay here I have derived
so del theta by del x is over here then plus V, so this is your V okay and then this is
your del theta by del y okay phi dash by root over of Re and in the right hand side conduction
side one by Pr then we are having del square theta by del capital y square so this is the
second derivative of the theta with respect to Y okay. So this equation we can further simplify and
cancelling different terms will be giving us phi f dash minus of f dash eta phi dash
plus eta f dash phi dash is minus f phi dash equals to one by Pr phi double dash okay.
So if you just do the multiplication all these terms will be coming okay eta f dash into
phi dash and f phi dash okay, so these two terms can be cancelled. So finally the equation will be coming as
phi dash plus Pr so Pr can be taken multiplication in this side so Pr into f phi dash minus phi
f dash is equals to zero okay and boundary condition as we have already derived that
phi dash zero will be equal to minus one and phi of infinity will be equals to zero okay.
So this is the equation and these two are the boundary conditions we have obtained okay
for constant heat flux cases. As we have taken this equation theta to phi relationship we
have taken in this fashion, so here let us try to see that how heat flux can be actually
link up with phi okay. So we know that phi already we have seen is
nothing but T minus T infinity by qw L by K so we say that till delta Te our non dimensional
term, our dimensional term for the scaling of the temperature okay. And in the right
hand side we are having the function of the phi as we have considered theta and phi equation
like this okay. So if you proceed further then phi can be written phi can be written
in terms of T minus T infinity by qw by K root over of 2 new x bar by u infinity okay.
So phi we have written in terms of temperature okay. So if you try to find out phi zero okay so
what happens at phi zero, so phi zero means at wall what is the profile of the phi okay
what is the value of the phi. So you can find out, so this will become actually Tw because
T at wall is Tw so Tw minusT infinity by qw by k root over of two neu x bar by u infinity
here why we are doing this because we want to find out actually what is the value of
Tw because in this case wherever we are having constant heat flux in this case the wall temperature
is unknown okay. So from here we will be trying to get what
is the value of this wall temperature right. So wall temperature if you do the side changing
and little bit of simplification of the previous equation so wall temperature can be written
in terms of T infinity plus qw by k root over of two neu x bar by u infinity into phi zero
okay. So main task here as qw is constant we have to find out what is the value of phi
okay at the wall once we can find out phi we can get the value of the wall temperature
okay. Let us also see what is the heat transfer
coefficient hx so hx is nothing but we know from adjacent layer which will be actually
taking the heat via conviction so qw is the wall heat transfer which is constant divided
by Tw equals to T infinity okay. So Tw we have already defined over here, so Tw can
be found out in this fashion so this is K by root over 2 neu x bar by u infinity into
phi zero okay. So Tw minusT infinity can be found out from here so this term will be coming
over here okay. So heat transfer coefficient we can obtain
in terms of phi only okay. If we proceed for that to obtain the nusselt number of heat
transfer coefficient so then we can find out once again nusselt number can be linked up
with phi over here because nusselt number is nothing but h xbar by K okay. So h already
we have found out and xbar and K if you link up with this one cancellation with this xbar
and xbar root xbar and this x bar will be giving you root xbar at the denominator and
then we can find out that this term is u infinity x bar by neu this will be giving you actually
Rex. So root over of Rex by root two into phi zero
any how this nusselt number is actually linked up with phi zero once again. So we have found
out if we can get this phi zero okay so all these wall temperature heat transfer coefficient
and nusselt number can be calculated. By the way to obtain this phi what we need
to do we need to actually do the solution of this equation along with the boundary conditions
okay. Solution of this equation is actually coupled with your Blasius equation f equation
which we have already showed in the previous lectures which is nothing but f tripple dash
plus ff double dash is equals to zero along with the boundary conditions okay. So these
equations are couple equations and needs to be solved numerically taking iterative solver. And if you do that and find out what is the
value of phi at wall that means phi zero then you can obtain this wall temperature heat
transfer coefficient and nusselt number respectively okay. Little bit of site change from this
nusselt number and Rex can give you Nux Rex to the power minus half is actually a sole
function of phi zero okay. Now here I will be introducing two different cases one is
for low Prandtl number limit another one is for high Prandtl number limit. So if you do
the solution coupled solution for this phi equation and f equation for a low Prandtl
number limit you will be finding out phi at wall and put in the equation of this nusselt
number. So then you will be finding out that this
is actually the correlation what we get for lower prandlt number, nusselt number Rex to
the power minus half is equals to zero point eight eight six prandtl number to the power
half okay, using your computational fluid dynamics knowledge if you just do the analysis
of these two coupled equation phi equation and f equation you can find out what is phi
zero for low prandtl number limit and this constant point eight eight six will be coming
from there okay. And similarly for the high prandtl number, prandtl number tends to infinity
it will become point four six three into prandtl number to the power one-third okay. Interestingly you can see that prandtl number
to the power half for low prandtl number and prandtl number to the power of one-third for
high prandtl number already we have shown in our constant temperature cases. So similar
dependence also we are finding out over here only the values are changing over here from
your constant temperature cases only because the phi value we need to obtain from activated
solvers okay. So this part I am not going but ultimately
I am showing you that these are the equations we can obtain for the nusselt number derivations
for two different limits okay. Next let us see that when we are having viscous
dissipation then how this equation is going to change okay. If you remain but till now
we have eliminated the viscous dissipation by considering very low Eckert number cases,
here this is very high Eckert number cases let us say we are having viscous dissipation
which cannot be neglected so this term will be remaining over there okay. In the previous lecture I have showed you
that phi star can be written as this one okay, so two del U del x whole square plus two del
V del y whole square plus Re to the power half del u del y plus Re to the power minus
half del V del x whole square okay. So here we have taken the transformation of small
y to capital Y in this fashion Y is equals to small y into Re to the power half and transformation
of small v to capital V we have taken in this fashion V is equals to small v into Re to
the power half okay. So here if we try to find out that which term is dominating once
again just like the previous one you will be finding out that this term is dominating,
because x is very high l is very high as compared to your deltas okay. V and Y these are of same order and here also
V is very small compared to X okay. So all these terms will be actually not dominating
with respect to this one okay. And if you take this Re to the power half outside with
whole square this will become Re and this Re and 1 by Re can be cancelled from the equation.
So ultimately Eckert number into del u by del capital y whole square becomes the dominant
term okay. All other small terms can be clubbed in this plus small terms okay. Now if you plug in this one in the equation
of your energy form in case of constant heat flux so the equation will be finding out is
changing in this fashion theta double dash plus prandtl number f theta dash is equals
to minus two Pr f double dash whole square so this extra term will be coming due to your
viscous dissipation term okay. And subsequently boundary conditions are like this theta dash
zero is zero and theta infinity is actually zero remember over here theta dash is coming
in case of constant wall temperature there was theta okay, because theta need to keep
for constant heat flux case okay. So heat flux is constant okay. Let me summarize what we have learnt in this
lecture first we have seen what is the choice of what temperature scale for non dimensionalization
of temperature. So unlike the previous one where we have seen the constant wall temperature
case the scale was Tw minus T infinity here we cannot take that also we have chosen intelligent
temperature scale like qw L by K where K is the thermal conductivity, qw is the constant
wall heat flux. And using this temperature scale we have showed
that force conviction of flat plate having constant heat flux can be reduced in this
form in terms of phi okay, so phi double dash plus Prandtl number of phi phi dash minus
phi f dash is equals to zero okay. We have also derived the corresponding boundary conditions
rather I can say that this boundary condition has been simplified from this temperature
scale non dimensionalization. So at eta equals to zero phi dash equals to minus one and at
eta tends to infinity phi tends to zero. So this is for updated value okay so this
becomes zero okay. We have also mentioned the nusselt number correlations for low prandtl
number limit and high prandtl number limits okay. So prandtl number tends to zero we have
shown that nusselt number into Rex to the power minus half is actually point eight eight
six prandtl number to the power half and for prandtl number tends to infinity we have shown
nusselt number into Reynolds number to the power minus half is equals to zero point six
three into prandtl number to the power one-third okay. And the end also shown that whenever
we take viscous dissipation into consideration in the energy equation over insulated plate
okay. (Refer Slide Time:23:22) By the by if you go back to the previous one
here I have not mentioned this is for the insulated plate okay, so till now we discussed
for constant heat flux here is the insulated plate so q w actually equals to zero. So in
that case find out the this equation we can come in this form which we also summarize
in the next slide. (Refer Slide Time:23:39) So theta double dash plus Prandtl number f
theta dash is equals to minus two prandtl number f double dash whole square okay and
the corresponding boundary conditions theta dash zero is actually zero and theta infinity
is tending toward zero okay. Let me also test your understanding in the end of this lecture
we are having three questions okay. (Refer Slide Time:24:04) In case of the forced convections over flat
plate having finite heat flux, choice of temperature scale has been made from? So choice of temperature
scale is very important so we are having four option difrence between the wall and the free
stream next one heat flux and thermal conductivity then free stream velocity and then none of
the above it takes some other temperature scaling okay. So obsivouly you have understood
because this is the thing in you lecture so correct answer is from qw and K qw by l okay. The second questions is the Nusselt number
for forced convection over flat plate with constant heat flux will depend on so we are
having four options Reynolds number, Prandtl number both Reynold number and Prandtl number
and none of these so remember this is for the constant heat flux over the flate plate
so obviosuly we know correct one is it will be depending on both the Reynold number as
well as the Prandtl number. Last question which statement is not correct in the context
of this lecture okay so here I have shown three equation over here first one is phi
double dash plus Prandtl number number into f phi dash minus phi f dash is equals to zero
the second one is theta double dash plus two f double dash square Prandtl number plus Prandtl
number f theta dash is equals to zero thired one is Nusselt number into Reynolds number
to the power minus half is equals to point four six three Prandtl number to the power
one-thired Prandtl number tends to infinity and all are not correct okay. Obvisuly you
can understand that all these equations are not correct so all are not correct is the
correct answer. So thank you, please visit our next lecture
which is about natural convection from uniform wall temperature plate and please keep on
posting in the enquiry of our discusssion for us thank you.

2 Replies to “Forced Convection over a Flat Plate: Uniform Heat Flux”

  1. As we choose partial theta / partial y = -1 for our convenience .. Is to possible to take any constant .. Kindly explain

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