Mod-01 Lec-24 Laminar External heat transfer with non uniform surface temperature

# Mod-01 Lec-24 Laminar External heat transfer with non uniform surface temperature

So today we will take up a problem on the
variable w temperature case and we will look at how to get the expression for the w heat
flux variation as well as the heat transfer coefficient so let us take the example of
a linear surface temperature variation that is of the form. T w (x)=a + BX okay so if you plot the surface
temperature as a function of X so at X is=0 that will be=constant an and from there
it varies linearly okay so this is the value T w at X is=0 which is=a so if you go
back to what we derived yesterday with the Duhamel superposition integral if you have
a continuous variation of the w temperature so we have derived the expression for the
w heat flux this was 0.33 1 x K / X P R 1 by 3 and you have T w at ?=0 – T 8 + 0 to
X 1 – ? by X the whole power 3 by 4 – 1 / 3 DT 1 by D ? x
okay ,so this is our expression you know now if for the limiting case where you do not
have any slope the slope is 0 so this is Steve all – T 8.
So this becomes=your flat plate with a uniform temperature expression okay so of course if
you have continuous variation but also intermediate temperature jumps okay so there you have to
introduce an additional d T x H you know for those temperature jumps in between okay so
apart from that this is the expression now we will substitute for the given temperature
profile whatever we require so for example in order to evaluate this Duhamel
integral we need to first get the slope of the w temperature profile and since this is
a linear profile it is very simple so in this case DT w by D ? will be for this particular
profile B right so we will substitute this x the equation. Let us call this as equation 1 so this is
T wet ?=0 is nothing but a okay so right at X is=0 this is nothing but a so I’m just
substituting for T wet ?=0 as a – T 8 + integral 0 to X 1 – ? / X to the power 3 by
4 the whole power – 1 / 3 DT w by D ? is nothing. But B so this will be B x D ? okay so now
all we need to do is evaluate this integral because everything else is known T 8 is given
for this problem whatever value it is known as is a constant B is a constant so we need
to find this particular integral and this particular integral whatever slope that you
get and put it here is of the following form which we will reduce it to so we can assume
a variable Z which is=? / X 3 / 4 okay. So I am just going to transform the variables
again so I am saying that ? by X 3 / 4 is=some other variable Z so therefore DZ by
D ? should be=3 / 4 ? /X 3 by 4 – 1 which is – 1 / 4 x with respect to D ? ? so X is
a constant so it will be 1 by X right so this is my derivative okay so therefore now I can
substitute for D ? in this integral in terms of DZ okay so this will give me my DZ is=3
by 4 x X power this is X 1 / 4 – 1 so this will be X -3 /4 okay x ? -1 by 4 x D ? alright
it is just I am just doing some algebraic manipulation there so therefore now I can
substitute for D ? I can transform these variables from D ? 2 DZ okay so I can substitute for
? / X T2 / 4 s Z and D ?. I can substitute in terms of DZ in fact I
can also write this in terms of Z ok now that is ? by X 3 by 4 so Z 1 by 3 I can put it
because this is already so my DZ will be D ? 4 by 3 here X 3 by 4 ? 1 /4 okay so ? by
X so I can write this as 1 by 4 X -3 by 4 I can take 1 by 4 X yeah so I can write this
as ? by X the whole power 1 by 4 this one – yeah right yeah right I can write it like
that because this is anyway 3 43 / 4 I can write it as X / X 1 /4 okay so now this is
nothing but ? by X to the power 1 by 4 is nothing but Z 1 / 3 okay so this I can rewrite
as that to the power this entire thing has that to the power one-third okay so what I
am doing is I am transforming all my variables from ? plane to Z plane okay.
So where ever I have Z that also how to include that so therefore if I substitute x this expression
0.33 1 a/ X P R 1 /3 Rd X to the power so you have a – T 8 + so I haven’t substituting
for D ? so 4 by 3 is constant 4 by 3 also B is a constant I can take out and also inside
the integral this is integrated with respect to now DZ therefore X also can be taken out
of the integral now 0 to this is originally 0 to X here so I can transform this to 0 to
one okay because that ?= X this becomes Z=one so therefore this will
be the upper integral will be one upper limit of integration so this will be 1 – ? 1 – Z
sorry 1 – Z – 1 by 3 and you have this Z 1 /3 here so that will also come outside and
by 3 x DZ okay just check 1 – Z to the power – 1 by 3 x that to the power 1 /3 DZ okay
so now I have anyway transformed that to this integral right here now how should I integrate
so I will just give you the formula. Now this is of the form of what is called
as a ß function okay the ß function can be expressed so generally the problems with
the Duhamel integral will be of the form of a ß function once you transform the variables
from ? to DZ okay so you will be ending up with a function something like this and you
can express this ß function. In terms of constants P and Q 0 to 1 Z P – 1
1 – Z Q – 1 valid okay this is valid for positive values of P and Q>8 okay so this is how
your ß function is defined okay, now if you compare this with this expression you can
see 1 – Z Q so therefore Q – 1 is=- 1 by 3 and P – 1 is=1 by 3. Right therefore Q
is=1 1 – 1 / 3 2 / 3 and P=1 + 1 / 3 4 / 3 okay so therefore this P and Q will be
nothing but 4 /3 , 2 / 3 okay, so therefore this entire term can be written in terms of
ß function. Okay so everything here as it is a – T 8 +
4 by 3 VX x ß function okay this entire integral is ß function the parameters are 4 / 3, 2
/ 3 okay so the integral is replaced by the ß. And why do we now need to write this in
terms of ß function because we can the ß functions are tabulated okay in fact very
specifically the ß function itself can be expressed as a function of another function
called the ? function okay so on ? function tables are quite common you know they are
tabulated for different values of the function. So therefore we will express the ß function
in terms of ? function as follows this is your ? function ß function we are generally
your ? function ? of say some variable s is represented as e power – X X S – 1 DX so this
is your ? function basically ok and this has been tabulated you know you can do this integral
numerically also for different values of s but this have been tabulated ? function charts
are there so you can look up for the values that we have here so therefore ß of 4 / 3
, 2 / 3 will be ? of 4 / 3 x ? of 2 / 3 / ? of what 6 /3 that is 2 okay so if you plug in
from the ? function tables which are available online.
You can Google ? function charts and you will find those nice charts for different values
of this you know for P and Q so if you plug in you will get this ß of 4 by 3 , 2 by 3
as 1 .2 0 8 7 okay so this is the resulting expression for ß 4 by 3 , 2 by 3 so if you
substitute that value x this so this is 1 .2 0 8 7 x 4 / 3. Which comes out as 1 .6 1 2 okay so now we
have a complete expression which gives you the variation of heat flux with respect to
X okay provided you know your constant A and B and you know the free stream temperature
T 8 all right so now therefore from this the heat transfer coefficient can be defined okay
so H of X can be defined as Q ? X /T w of X – T 8 okay now T w of X – T 8 can be written
as a + B X – T 8 so therefore you have this entire 0.33 one a by X B R to the power one-third
6 1 2 BX the entire thing / T w – T 8 which is nothing but a + BX – T 8 okay so therefore
for the given constants you can now determine the local variation in the heat transfer coefficient
and for the limiting case. Where your B=0 B=0 gives me a uniform
w temperature okay so for that case how does it reduce this becomes a – T 8 here the denominator
this cancels with a – T 8 here B is anyway 0 okay so then you will what will be the expression
the constant w temperature that is 0.33 1 K by X P R 1 / 3 re X 1/2 okay so this is
your constant w temperature boundary condition so for the limiting case where B=0 you retrieve
your constant w temperature heat transfer coefficient okay so it is a very straightforward
method as such you know so what it finally means that so you can also solve this by dividing. These x piecewise constants like we did yesterday
you know you can assume that you can represent this by piecewise constant like. This you right you can divide this x piecewise
constants and instead of the integration that we did here we will replace this by discrete
summation okay so if you do that you can also get somewhat similar expression but that will
be in a slightly discrete form so in that case you will have something like point 3
3 1 x K by x re X power 1/2 P R power one-third so this will be e 1 0 – T 8 so when you differentiate
it the original profile you have a fee so that will be – K DV / dy at y=0 this will
be nothing but the heat transfer coefficient H okay.
So that H you have already substituted as 0.33 1k by x re X power 1/2 PR power 1 by
third x 1 – ? by X the whole power – 3 by 2 GJ so what is that can you go back and revise
the expression for H of X 0.33 1k by x re X 1/2 P R power 1 third 1 – ? by X to the
power what 3 by 4 the whole power – – 1 / 3 okay so this we had to substitute x the expression
for Q all of X which was so you had the original expression for T w – T 8 that is – T 8 was
T w 0 – T 8 x fee of 0 , X , Y + summation of n=1 to capital n d T w and fee of X , Y
so this was what we saw a study this was for the local variation in temperature.
We have super post the solutions where you have a uniform temperature okay and one by
one so then the incremental temperatures so all of that when we superpose so first you
have this is the basic solution that you have + the incremental solutions which is basically
this okay so now when we want to calculate the heat flux so we had to say – K x D ? / dy
at y=0 which was nothing but the heat transfer coefficient H so this we had to substitute
x this expression so therefore this is taken out as common you have T w 0 – T 8 as one
of the terms and the second term will be the rest of the terms will be in n=1 to capital
n number of discreet intervals you have d T w and okay x 1 – ? by X (3) / 4 full power
– 1/ 3 okay. So this is what you will have if you have
a discrete variation if you had a continuous variation you will replace this by an integral
okay integral over D ? okay now you have a discrete variation therefore you just substitute
for d T as it is and of course the heat transfer coefficient for the first location where ?=0
that is no unheated starting link so therefore you don’t have this term for that for the
subsequent boundaries and conditions you maintain at ?=? 1 ?=? 2 so there you have unheated
starting link. So there you have to substitute the values
of so this will be corresponding value of ? ? n okay so by doing this also you can calculate
your local wall flux variation instead of using the do Hamill’s integral method you
can just linearly superpose this is the superposition method right so you can divide this continuous
curve x small discreet intervals where you have no variation of temperature like this
you can substitute that x this discreetly and you can also estimate the w heat flux
okay, so both will be more or less the same they are continuous is the more accurate because
you are taking x account the slope accurately okay so this is to just give an example okay
how you take a problem. Where suppose you have a w temperature variation
like this and you can use either the Duhamel integral or a simple superposition technique
and you can calculate your local heat transfer coefficient and your w heat flux okay so any
questions on this okay now for more complex profiles you know you it’s more likely that
most of the ball temperature variation cannot be approximated just by a straight line it
will be more complicated profile so for a more complicated w temperature variation what
is common practice is that we can approximate the w temperature variation as something like
power law series so instead of using an A + B X relationship we can write this as a
+ summation n=1 to N B and X okay. So this is the power series expansion
so which is which is K which can be used to approximate more complicated nature of profiles
you know if you have a profile something like this so you can you can use power series expansion
you can fit the coefficients to know you can do a regression fit the coefficients such
that you can approximate this curve with the power series expansion okay so this is a better
way to represent this than using a straight line right so you can substitute this now
x this expression here so you had a term here DT w by D ? so now you have to calculate DT
w / D ? for this so what will be the expression for DT w by D ? summation n=1 to n you have
n B and X power n – 1 or ? power and – 1 okay. So this has to be now substituted x the expression
where we had DT by DT w by D ? okay and if you do that the resulting expression for the
heat flux comes out to be everything is the same only you have the summation term 3 3
1k /X so everything up to here is the same except you have the summation term and everything
inside the summation term goes there so n=1 to n you have n be n so now so this will
be X n – 1 x there will be an X which will come out of the transformation okay so that
will be giving you X n okay x the other the ß function will be there as it is so you
will have ß now this ß function also will become a function of n okay so where this
ß function we can be written as ? function of 4 by 3 originally it was 4 by 3 now it
becomes 4 by 3 n x ? function of 2 by 3 / ? function of 4 by 3 n + 2 by 3 okay so now depending
on the value of n that you use ok so the value of ß will change and then you have to sum
them over all the values of n. So suppose you are using five terms you have
to sum them for each value of n and for all the 5 times you have to sum them together
so then that that will give you the variation if you have a more complicated profile you
know you can approximate that with the power series expansion and you can use this expression
to calculate the local heat flux variation okay now the question is given local variation
in the temperature profile we can use this to calculate the local heat flux but what
about the other way suppose your w boundary condition is a locally varying w heat flux
okay so how do you calculate the local w temperature as well as the heat transfer coefficient okay
so that is also a little bit more complicated derivation.
I am not going to do that I will just only give you the final solution for the for the
w temperature variations. Okay for a non-uniform w flux so for this
case you can calculate the w temperature variation can express it as follows .6 to three by K
P R – 1 / 3 re X – 1/2 you have 0 to X 1 – ? by X 3 / 4 the whole raised to the power – 2/3
x Q okay so this is the local variation in the w flux whatever you have so that can be
substituted x this and you can get the corresponding w temperature variation okay so this is this
given in your textbook case in Crawford okay he has also not derived it.
But I think there is a reference to some literature where they have done it and they have shown
that shown this kind of an expression anyway so this is beyond the scope of your this thing
but you should understand that you can do either of these in when using the approximate
solutions given non-uniform w temperature how do you calculate the variation in the
w heat flux or given a non-uniform w heat flux how can you calculate the variation in
the w temperature so both are possible by using the approximate methods so I think with
that we will kind of wrap up the external laminar external flows external boundary layer
flows. So we have covered quite a bit you know we
have got almost covered whatever possible similarity solutions under external boundary
layers and also the approximate methods we know whatever external laminar similarity
solutions they have a complimentary integral methods also integral solutions also like
we have seen if you have the Falkner Skan solutions for which problem you have similar
foreign Carman pool house and solution when you use the integral methods okay of course
like the blusher solution there you have you can approximate some velocity profile and
very easily find out the expression for say nacelle local nusselt number okay.
So whatever is possible in fact you can also use the approximate solution for example a
flow with transpiration you have boundary where you have porous boundary with suction
or blowing so we have derived the similarity solution for that in fact we have we have
identified the condition for the variation of the suction profile velocity profile so
that you can get a similarity solution okay so the same way we can derive an approximate
by using approximate method you can we can derive expressions for the local skin friction
coefficient as well as the nusselt number variation with transpiration so every similarity
solution has a counterpart. In the approximate methods and more than that
you can also derive some special cases such as the unheated starting length which you
cannot derive by similarity solutions and also cases such as these where you have non-uniform
w temperature variation of any of any given profile which you can approximate as a power
series expansion or a non uniform w heat flux okay so for these kind of boundary conditions
you know it will the similarity solutions are not possible or it becomes very rigorous
so their approximate solutions are much easier okay so this is in a nutshell giving you an
idea what we covered so from the next week onwards.
We will look at internal laminar internal boundary layer flows okay so they are actually
strictly speaking the boundary layer concept doesn’t have a meaning that the way that laminar
external flows has okay the strict definition of boundary layer flow does not hold for internal
flows okay because once you have a fully developed flow both the boundary layers merge and everywhere
you have viscous effects there is no place where you can use potential flow and approximate
that with the potential theory and somewhere you can solve with the solving the full navier-stokes
okay so therefore we have to resort to a complete solution of the navier stokes equations in
some cases. In some cases we can make approximation to
the velocity gradients or the temperature gradients okay so there we can obtain exact
solutions of reduced form of the navier-stokes equations okay so that is also very important
and interesting because most of your practical problems in heat transfer okay although there
are many external flow problems you will find most of the heat exchangers they are encountered
you will be encountering internal flows there and in those cases you should understand the
approximations that you can make and how you can get the expressions for local variation
in the nusselt number and in a fully developed case how the nusselt number variation there
is no variation in the nusselt number and so on okay.
So that we will cover in the next 9 to 10 classes starting from next weeks in about
three weeks I think we should be able to cover the laminar internal flows and then from the
following we converts that is about the fourth week of March professor Koehler will start
turbulent flows so that will be for about seven lectures or so and or seven or eight
lectures and maybe natural convection for another seven or eight lectures you.
Laminar External heat transfer with non uniform surface temperature
End of Lecture 24 Next: Laminar internal Forced convection -Fundamentals