# 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”

Plz solve the coupled equations to derive the expression of Nusselt number

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