Natural Convection: Uniform Wall Temperature

# Natural Convection: Uniform Wall Temperature

Hello, let us welcome you all in our eight
lecture of convective heat transfer course, in this lecture we will be discussing about
natural conviction, in earlier versions of this course what we have seen that there will
be a flow over the flat plate but in this case we will be seeing there is no flow over
a hot plate rather the flow is being created by virtue of the temperature gradient between
the plate and the free stream. So this type of convection is actually called
natural conviction and in this natural convection we will be actually concentrating in this
lecture on uniform wall temperature, so today’s lecture is based on natural convection uniform
wall temperature, okay. So let me tell you that what things will be covering in this
lecture at first I will be introducing you what is natural convection, how its velocity
and temperature profile looks like, okay. Whenever you are having a hot plate in vertical
orientation, remember for natural convection vertical orientation is utmost important in
the previous lectures you have found out the plate was kept horizontal and there was a
flow which is actually force conviction, in this case the plate will be vertical and natural
conviction will be occurring so I will be introducing you what is the velocity and temperature
profile. That too will be considering the plate is
being kept at constant temperature, okay. As I have told you that there is no concept
of forced velocity so what will be the velocity scale determination of that velocity scale
is very important, so I will be introducing you how velocity scale for natural convection
can be considered especially for constant temperature case, constant temperature of
the wall case, okay. We will be deriving the momentum and energy equation for flow around
a vertical hot plate which is kept at constant temperature, okay. And at the end you will be mentioning sets
of equation and Nusselt number correlations for both high and prandtl number fluid flow
around the vertical hot plate. So let me first introduce you with the situation
what happens in case of natural conviction as I have told you that in case of natural
convection there will be no predominate flow, flow will be rather generated by virtue of
convection so let me show you this is the plate whatever I am considering over here
is kept in the vertical orientation so g is actually perpendicularly g is actually coming
from top to bottom. So in the downward direction g is acting over
here gravity is acting over here and as we are interested in uniform wall temperature
case, so we are keeping that the wall is having at constant temperature TW, okay and let us
consider that the fluid which is there around is at in temperature T infinity and there
is no flow predominant flow in the fluid so a fluid is at rest away from the plate, okay. So free stream velocity is zero, okay. Now
as we are having high temperature TW over here in comparison T infinity which is a free
stream temperature will be finding out there will be some temperature profile because at
wall you are having very high temperature and away from the wall you are having low
temperature, so the temperature profile you will be reducing in such in this fashion.
So here I have shown you the temperature profile, and what we can consider wherever the temperature
has reached to ninety nine percent of the free stream temperature we can call that point
as thickness of the boundary layer, okay like this if we consider along the plate axial
direction that means x bar so we can find out locus of points okay which actually resembles
ninety nine percent of the free stream temperature. So if we join those we will be having the
thermal boundary layers so this actually is the thermal boundary layer, okay. On the other
hand to give this temperature profile we will be finding out there is some flow around the
plate, okay though away from the plate we are having zero velocity okay and at adjacent
to the plate due to your no slip boundary condition we will be having zero velocity. But around the plate you can find out that
to enhance the convection from wall to the freeze team you will be having some velocity
so we are considering that at wall we are having zero velocity and far away from the
wall once again we have regained the zero velocity and in-between we are having a velocity
profile like this, okay. We are not saying that this velocity profile will be symmetric
or it can be you know taking high value very near the wall and then slowly going as towards
the zero velocity, okay. So just like your concept of velocity bound
layer in case of forced convection here also velocity boundary can be constructed but in
this case as there is zero velocity in the free stream team so we have to reach something
around zero, ninety nine percent of the zero velocity so those points if we find out along
the axial direction axial location of the plate then and then join those points will
be finding out the velocity boundary layer. So this inner curve is actually our velocity
boundary layer and outer curve is our temperature or thermal boundary layer, okay. But here
I have shown you only the representative case quickly I will be showing you what happens
in case of you know two different types of extend of fluid high prandtl number and loop
prandtl number of fluids, where you will be finding out there is contest between the velocity
boundary layer and the thermal boundary layer. Sometimes velocity boundary layer it is becoming
dominant and sometime thermal boundary layer becomes dominant okay, so this is one representative
where I have shown velocity boundary layer thickness delta is smaller than thermal boundary
layer thickness delta t but reverse cases also possible, okay. Quickly we will be coming
to those situations by the way let us then quickly try to understand that what are the
governing equations involved with this one. So first we will be introducing from fluid
mechanics what is the continuity equation, okay. We are considering that we are having
incompressible fluid over her around this vertical plate so we are actually as neglecting
the density changes so if you consider that one then continuity equation simplifies to
del u del x plus del u del y is equals to zero okay. Then if you see the momentum equation
let us consider two dimensional plates over here in x and y coordinates x-bar and y-bar
coordinator rather. So we will be finding out that our x momentum
equation it goes like this so this is your inertia term pressure gradient term, viscous
term and subsequently this term will be there due to bouncy okay now as the plate is actually
aligned with the x- axis so the whole term of the buoyancy will be coming in the x momentum
equation there will be no representation of the buoyancy in the y momentum equation. Y momentum equation is very simple in terms
of v bar so this is the inertia pressure gradient in terms of subsequently the viscous terms,
okay. So after this continuity and momentum equation as we are in convective heat transfer
so energy equation will be important so let us see the energy equation, so energy equation
you see in the left hand side we are having as usual convection and right hand side we
are having as usual conduction term. They are linked up with the alpha which is
nothing but your diffusivity, okay thermal diffusivity then as we have done earlier let
us try to non dimensionalize this equations by taking different scales, so first for the
locations x-bar and y-bar will be non dimensionalizing with L which is the characteristic length
scale or you can length of the plate okay and we actually construct x and y as non dimensionalize
parameters, non dimensionalize lengths scale, okay. In plate direction and cross plate direction,
okay. For temperature let us consider theta, now we have mention the temperature of the
plate is TW and free stream temperature is T infinity so with the help of this two we
are non dimensionalize and temperature T so T – T infinity by TW – T infinity this
we are considering as theta, okay. So as a result near the wall theta will become one
and away from the wall theta will become theta will tend towards zero, okay. For velocity we are considering some characteristic
velocity scale U not okay this will be also U not so some characteristic velocity scale
U0 we are considering so U and V these are U- bar/ U not and V equals to V bar by U not
okay, pressure we are characterizing by P-bar by rho u square so u square is having actually
a unit operation so p-bar by u square is actually capital P which is non dimensionalize pressure,
so if you take all this non dimensionalization schemes and try to modify this continuity
momentum and energy equations. Our system of equations will be changing,
so let us try to see from there what this equations will be taking the form but before
that let us see what is this U not because U not is some fictitious quantity what we
have taken okay so to take this U not what we are considering that what is the order
of your inertia term so if you see the x momentum equation the inertia term will be becoming
U not square by L okay. So this is U not into U not and here it is
x so that means L, so U not square by L is the dominant term over there, so what we are
considering U not square by L which is the inertia term order is equals to the last term
that means the buoyancy term over here g beta T – T infinity, okay. So if you consider
that so g beta (T – T infinity) is equals to U not square by L so here actually we have
equated the inertia order along with the buoyancy order if you do so, so you can get the idea
what will be the characteristic velocity U not. So U not will be root g beta L Tw minus T
infinity okay, next let us try to find out that what will be the what will be the coefficient
in our viscous term so if you see the viscous term in the previous equation I will show
you in the viscous term we will be having gamma and then which is kinematic viscosity
from there here we will be getting U not and here you will be getting L square and from
this side we will be getting U not square by L so if you divide the whole equation by
U zero square by L the coefficient in front of the vicious term will become gamma U not
square by L and U not square by L okay. So after reduction you will be getting this
is nothing but gamma by U not L okay, so as we have already found out what is U not we
can straight away put this U not value over here to get what is the coefficient of the
viscous term, so if you see the coefficient after cancellation and putting this value
of U zero it is becoming gamma not square divided by g beta L cube) TW – T infinity.
Now this by g beta L cube (TW – T infinity) by gamma not square is actually special quantity
or special non dimensional number we call that one as actually Grashof number, okay. So this is kinematic viscosity so this is
Grashof number so we are getting this coefficient of the viscous term is becoming one by root
over of Grashof number okay, so Grashof number gives the physical significance that what
the effect of buoyancy and what is the effect of viscosity, okay. So that is actually non
dimensional number, so once if we see that what our equations are taking the form after
non dimensionalization so you will be getting inertia term is having simplified U del u
del x plus V del u by del y because we have divided with the coefficient whatever we had
over there throughout so we have got one by root over of Gr in front of the viscous term
and as we have equated the buoyancy and inertia force so there is no coefficient in front
of the buoyancy force. Because we have divided with the coefficient
of the buoyancy term so coefficient of the inertia term so you are finding out that here
it is throughout divided so it is becoming one okay, why momentum equation similarly
you too will be getting U del v del x plus V del v del y is equals to minus del p del
y plus one by root over of Gr del square v by del x square okay, here theta term will
not be there as plate is vertically oriented, okay. Similarly for energy equation you will
be finding out along with one by root Gr we will be having Prandtl number also over here. Because in this case in front of energy equation
in place of gamma we had alpha so to bring the effect of the Grashof number what we can
do, we can write down alpha by gamma into gamma okay so that gamma will be giving the
effect of the Grashof term but an alpha by gamma will become the Prandtl number, so here
you can see root over of Gr into Prandtl number is coming over here, okay. Next let us try to do some sort of scale analysis
from all these three equations whatever we have derived, so let us start from continuity
equation now we know that what we can do, for the along the direction of the plate we
can assume that x which is along the direction of the plate is of the order of one and for
the cross direction or the vertical or the horizontal direction we can call y is of the
order of the del not, del not is actually the some thickness. Whether it can be velocity boundary layer
thickness or thermal boundary layer thickness that comes later on whenever you are considering
different prandtl number, now from continuity we can straight away write down that du, del
u del x need to be in order of one so if x is order of one then obviously u will be order
of one and in case of Y if it is order del not , del v del y needs to be in order of
one. So V becomes order of del not, okay. Now let us see the viscous term if you see
the viscous term what is the order of the viscous term let me show you the viscous term
once again, so if you see the viscous term once again over here one by root over of Gr
then we are having del square u del x square plus del square u del y square okay, so in
this case you can find out that the order will be coming will be finding out that this
term is becoming dominant, okay. Because Y is of order del not. So this term will becoming dominant but x
is of order one so this term is becoming dominant and order will become one by root over of
Gr into one by del square, okay so this is the order of viscous term one by root over
of Gr int one by del not square okay. In a similar fashion if you try to see from the
inertia term so in case of inertia you will be finding out the order as we are having
V and Y over here, so this will be cancelling out okay, so delta not and delta not will
be cancelling out. U is of order one so this becomes order one
similarly all this quantity U and X these are of order one so the total inertia term
is becoming of order one, so inertia term is having order one, okay. Whereas the viscous
term is having this order right. So if we equate the inertia and viscous term to get
the momentum equation back, so we get one by root over of Gr del square is of order
one and from here we get del not is actually Gr to the power minus one by four okay. So we have got the thickness of the boundary
layer is of the order of Grashof number to the power minus one by four. So let us take
some stretching terms formation, so from small y as we have already known that what is the
order in the y direction we can take y into Grashof number to the power one-fourth is
actually capital Y and similarly to satisfy the continuity we need to take for small v
also, so small v into Grashof of number to the power one by four is actually capital
V, okay. So if we use this stretching transformation
our continuity equation will not be changing because in the first term there is no capital
Y or capital V, in the second term we are having capital V and capital Y but this Grashof
number to the power one by four will be cancelling from denominator and numerator okay, so continuity
equation will not change, in x momentum equation we can find out in inertia term there is no
change. As it is of order one already we have told,
okay. In case of your viscous term you can find out that your cross wise viscosity that
means del square u del y square is actually releasing one Gr to the power half so that
Gr to the power half and its coefficient are earlier coefficient one by root over of Gr
is cancelling out and there is no coefficient in front of the del square u del y square
term but del square u del x square will be keeping the coefficient one by root over of
Gr, okay buoyancy remains as usual as theta and pressure gradient remains as minus del
p del x, okay. As we have not done any transformation for x so this term will be remaining same
so this is my x momentum equation. In a similar fashion Y momentum equation can
be constructed so this is Y momentum equation so as we are having over here V so what we
have done the coefficient of V actually has taken this one Gr to the power minus one by
four similarly over here we are having two V’s and one Y so ultimately Gr to the power
minus one by four will be coming out, so that we have taken common over here, in the right
hand side for the pressure gradient term we will be having Gr to the power one by four
okay because Y is over here. For the viscous term so for the first term
del square v del x square there will be only V over here so Gr to the power minus one by
four will be coming out and earlier we are having Gr to the power minus half so that
will be giving rise Gr to the power minus three by four and for the last term del square
v del y square so we will be having one V at the bottom sides so that along with one
by root over of Gr will be giving rise to the Gr to the power minus one by four okay.
Subsequent changes of side will be giving you del p del y is equals to something in
the right hand side which is having everywhere Grashof number, okay. So this is the final
form of the momentum equation we can write down, okay. Similarly energy equation so you see left
hand side will be having no change because this is your convection side having order
one so U and X those are of order one, V and Y they are having stretching’s but Grashof
number is releasing in denominator and numerator those will be cancelling out, in the right
hand side conduction term you see for del square theta del y square if we convert that
into capital Y, capital del y square. So one Grashof number to the power half will
be releasing so that Grashof number to the power half and one by root over of Gr will
be cancelling out and leaving out this one by Pr into del square theta del y square but
same thing will not happen for del square theta del x square so you will be finding
out one by root over of Gr Pr okay del square theta del x square right, so these are the
equations form equations of different forms after this stretching term formation. Now
let us take the limit of Grashof number tends to infinity. So this Grashof number tends to infinity comes
out whenever you are having significant amount of temperature drop between the wall vertical
wall and the free stream compared to the you know viscous effects, okay. So here we take
this Gr number tends to infinity if you take Grashof number tends to infinity in all the
equations wherever you are having Grashof number to the power minus power. So those terms can be cancelled, so you see
from x momentum equation last term can be cancelled okay and from Y momentum equation
whole right hand side can be cancelled okay everywhere we are having minus power okay
of the gross of number and from the energy equation this delta square theta del x square
can be cancelled, so ultimately the equations comes out to it continuity in this fashion
x momentum in this fashion this is your Y momentum equation, simply del p by del y is
equals to zero and finally this is my energy equation which is important for this course
okay. So delta square theta del x square is absent over here, right. So let us take further simplicity let us consider
that from here we can get del p del y equals to zero so that means P is a function of x. And let us take that P which is function of
x is actually a constant okay, so if we take a constant then obviously in x moment equation
minus del p del x is equals can be dropped down okay, so here you see from x moment and
equation del p del x we have dropped down to make our equations simpler in look okay.
So this is my x momentum equation, y moment term equation we have not shown over here
because this derivation now we have actually taken from y momentum equation. This is my energy equation as usual whatever
we have derived in the last slide, let me tell you the boundary conditions also so at
Y equals to zero which is nothing but adjacent to the wall okay, adjacent to the wall obviously
no sleep and no penetration will be giving me u equals to zero, V equals to zero. And
from the non-dimensionalization of the theta we will be finding out that at the wall temperature
is Tw so this is becoming Tw minus t infinity by Tw minus T infinity which is equal to equals
to one, okay. And away from the wall so Y tends to infinity,
capital Y tends to infinity, we will be finding out there is no velocity so u equals to zero
okay, and theta will be becoming zero because t infinity minus t infinity by t w minus t
infinity, is the theta which is nothing but zero okay, so these are the equations and
boundary condition subsequently. Let us try to modify this things using our stream function
concept, all of us know that u is nothing but del sai del y and V equals to minus del
sai del x where sai is the stream function, okay from fluid mechanics we know this.
Let us try to get that what will be the equation forms involving this stream function, so obviously
continuity equation will not be giving us something so I am not writing continuity equation
over here, basically this things have been derived from the continuity equation only
so let us see the momentum equation x momentum equation so if you plug in this u over here
and V over here then we will be writing down del sai del y into del square sai del x del
y. So this is nothing but your del u del x minus
del sai by del x this is nothing but your V and del square sai by del y square this
is nothing but your del u by del y. On the right hand side we are having del square u
del y square is nothing but del cube sai by del y cube plus your by in theta miss giving
theta. And in energy equation first term is actually your u del sai by del y is actually
your u and del theta by del x is remaining like this for V we are writing minus del sai
del y and this is your del theta by del y which is coming from the second term of the
convection. In the right hand side we are having one by
Pr del square theta del y square okay as usual from the energy equation. Boundary condition
will not be changing much, so at Y equals to zero, U equals to zero means del sai del
y equals to zero, V equals to zero means from here we can get if we integrate once we will
be finding out sai equals to zero and energy equation temperature boundary condition is
not changing theta equals to one. And for away from the wall Y tends to infinity we
are having del sai del y tends to zero and theta tends to zero, okay. So this is your
u tends to zero and theta tends to zero right. Let us do now little bit further try to have
some stretching transformation from this stream function equations, let us define that X star
is equals to e to the power alpha one small okay, so we are taking this exponential relationship
where Alpha1 is one constant, okay. Let us do similar things for Y star, sai star and
theta star okay, so Y star is equals to e to the power alpha two capital, sai star is
equals to e to the power alpha three sai and Theta star is equals to e to the power alpha
one theta, this we are doing for finding out the similarity variable in terms of x and
Y, okay. So here we are having four unknown alpha one,
alpha two, alpha three and alpha four we will putting back this stretching variables in
your vowelling equations to find out the similarity variable. So if we put all these stretching
transformation in our equation and convert all those without star terms to star terms
we will be having one co-efficient of e to the power alpha one plus two alpha two minus
two alpha three from our inertia term in x momentum equation. And in viscous term we
will be having e to the power alpha three a two minus alpha three okay, as we are having
Y cube so three alpha two and as we are having sai it is becoming alpha three okay, alpha
three and alpha two like this, okay. And for the theta term we are having e to
the power minus alpha four okay, similar thing can be done for the convection side so our
equation is becoming of e to the power alpha one plus two alpha two minus two alpha three
minus alpha four okay, so sai, theta and Y and x so all alpha one, two alpha two, two
alpha three and alpha four will be coming into picture, and in the conduction side we
are having e to the power two alpha two minus alpha four okay, so theta is giving rise alpha
four and Y square is giving rise to two alpha two, right. And from the boundary condition
so Y equals to zero means, Y star equals to zero okay, and Psi star equals to zero means
Psi equals to zero but here the second uequals to zero, V-zero boundary condition gives rise
del sai sorry, u equals to zero boundary condition gives del sai del y e to the power alpha two
minus alpha three equals to zero, okay. So and your temperature boundary condition
gives e to the power minus alpha four theta equals to one okay, in a similar fashion away
from the wall Y star tends to Infinity we can find out del sai star del y star e to
the power alpha two minus two into alpha three tends to zero okay, so this is nothing but
your u tends to zero and e to the power minus alpha four theta tends to zero this is nothing
but your theta tends to zero, okay. So now from here all these things if we equate the
order from both the sides we would like to find out what is the relationship between
alpha one, two alpha two, two alpha three and alpha four okay. So from the boundary conditions, first boundary
conditions clearly we can see that alpha four leads to be zero otherwise theta star cannot
be equals to one okay, because theta equal to one or the boundary conditions otherwise
we will be making it complicated. So definitely needs to be equals to zero okay, as we need
to have this one equals to one, okay. And then from here we can see that this three
alpha two minus three alpha three definitely needs to be minus three four okay, as we have
already prove that alpha four equals to zero so three alpha two minus alpha three also
needs to be zero, so three alpha two minus alpha three needs to be zero, okay. So from here we are getting alpha three equals
to three alpha two okay, in a similar fashion if we equate the inertia and viscous terms
co-efficient then we can find out relation between alpha one and alpha two in this fashion,
so alpha one plus two alpha two minus three alpha three needs to be equals to zero, because
inertia and buoyancy terms these two needs to be of say molder okay, so here you can
find out we are getting Alpha1 in terms of alpha two as alpha one equals to four alpha
two okay. So here we have obtained X eight equals to e to the power four alpha two into
x, Y star equals to e to the power alpha two into y, sai star equals to e to the power
three alpha two into sai and theta star equals to theta as alpha four equals to zero okay. Now let us try to consider that e to the power
alpha two equals to c, so ultimately we will be getting x star equals to c to the power
four, y star equals to c y, sai star equals to c cube and theta star equals to theta.
Now from here let us try to construct the similarity variable, so we can easily write
down from this two equation y star by x star to the power one by four so y star by x star
to the power one by four will become small y divided by capital y to the power one by
four okay, so here you see c and c to the power four by one by four is actually cancelling
out it side, so this becomes the similarity variable, okay. On the other hand if you see the third equation
and the first equation here from also we can write down sai star by x star to the power
three by four is equals to sai by x to the power three by four okay. So here also we
can get that what will be the functional dependence between the sai and X and as usual theta star
equals to theta will be remaining over there. Now as we have got the idea that what can
be my similarity variable, so let us write down this similarity variable as y by x to
the power one by four okay, so I am writing over here eta which is a similarity variable
y by x to the power one by four and as I do not know what is the constant term will be
coming in front of that let me keep that one as A. So I am defining eta equals to A into y divided
by x to the power one fourth, in the same fashion let us define sai as a function of
eta so what we are writing sai is actually x to the power three by four from here we
can get that one, x to the power three by four into F okay and as we do not know what
will be the constant we are keeping the constant as beta okay. And theta always from this one
we can write down theta can be expressed as a function of eta, so theta is actually theta
of eta, right. So now our task simplifies to finding out the values of A and B okay,
to explicitly find out the similarity variable and stream function dependence over eta. So let us do that so in order to do that first
we need to find out the derivatives of different terms and put in the equations, so first let
us find out what is the eta’s derivatives so similarity variables derivative, so del
del x of eta is minus eta by four x and del del y of eta is A by x to the power one by
four okay, and let us side by side do del theta by del x also, so del theta by del x
will be important for your energy equation, so this becomes del theta by del eta into
del eta by del x okay, so del eta by del x I can pick up from here and del theta by del
x will become theta dash, okay. So minus eta by four x theta dash in a similar
fashion del theta by del y we can write down del del eta of theta into del del y of eta
okay, so these becomes theta dash and this del theta by del x is nothing but A by x to
the power one by four so this is the del theta by del y coming out to be. So let us now put
everything in, at first everything let us put in no, before that we need the value of
del sai by del x also so this is del sai by del x okay, so for finding out del sai by
del x what we are doing the previous equation whatever we have over here we are making the
derivative with respect to x, so this is also function of x and F is also a function of
x so we are having derivative of two multipliers. So we are using that multiplier rule, so here
you see first we have kept f constant and we have made the derivative of x to the power
minus three by four okay, x to the power three by four and here we have kept x to the power
three by four constant and here we have made the derivative of F, as F is the function
of eta so d eta of dx also comes into picture using the value of eta so d eta of dx we can
simply this and it will be coming as B by four x rest to the power one fourth into three
F minus eta f dash okay. In a similar fashion del sai by del y can be calculated this is
simple, so B into x rest to the power three fourth as x is not a function of Y so this
will not be making derivative this time. So only F dash is having and A by x to the
power one fourth is coming due to B eta by dy, okay, so simplified form is AB x to the
power half into F dash. second derivative of sai with respect to Y you will giving AB
x to the power half F double dash into A by x to the power one by four okay, so ultimately
it is becoming A square B x to the power one fourth F double dash. Third derivative of
sai which is also coming in our viscous term so this is, this will be becoming A cube B
F triple dash okay, and cross derivative of sai del square sai del x del y will be, this
you can achieve by making derivative with respect to x of this term or derivative of
this term with respect to Y, okay. So both will be giving you same result and
final simplified answer you will be getting as AB by four x into two F dash minus eta
f double dash so all this derivatives we have obtained for sai okay, and theta also we have
obtained only term left is that del square cube so all this terms you can get in this
fashion. So let us put everything in our momentum equation
so this we have put in the momentum equation, so you can find out this is your u okay so
this is your u del u del x okay, and this one is your V so this one is your V okay,
and this part is u actually your del u del y okay, and in this side we are having over
here del square u del y square, del square u del y square and theta term will remains
like this okay, in the momentum equation. Further simplification of this one u will
be giving you in this fashion the equation will come out in this fashion okay, where
this two term can be cancelled okay, so if you do the simplification this is our equation
involving A and B. Remember here also still A and B prevails. From your energy equation
if you do the same things so this is your u, this is del theta del x, this is my V and
this is my del theta del y and in this side we are having del square theta del y square
into one by Prandtl number. Simplification of this one if you do for the
simplification this will come further one steps simplification if you do then it will
be coming like this and at the end you can find out this term and this term can be cancelled,
so theta double dash plus three Prandtl number B by four A F theta dash is equals to zero
becomes an energy equation. So this is the momentum equation this is the energy equation,
but still here A and B terms remains so let us try to find out A and B. So first let us see that what is happening
over here, here in both this cases AB by four is the co-efficient and here we are having
A square by Pr. So first let us try to see from B what we
can say it is that B by four A equals to one, so B by four A you see here in the energy
equation we are having B by four A if we can make this one this equation look like a simplified
one, so first let us try to make B by four A equals to one and also try to make that
A cube B is actually equals to one, A cube B is here so A cube B is equals to one, so
if we make like this then from this two equations we can get the value of A and B by replacing
A from here to here we can get the value of A and B and that comes out to be one by quadruple
root of and B it will becomes out to be four to the power three by four. Once we know the value of A and B getting
the similarity variable is not difficult so similarity variable becomes Y by four x to
the power one by four and sai becomes four x to the power three four into F okay. so
once again if you give back all this A and B values in your equations whatever we have
shown over here this two equations so my final form of equation comes out to be F triple
dash plus three F F double dash minus two F dash square plus theta is equals to zero
this is the momentum equation and energy equation comes out to be theta double dash plus three
Pr F theta dash equals to zero okay, so this and this we have made actually one okay and
boundary conditions that eta equals to zero, F equals to zero, F dash equals to zero, theta
equals to zero okay, and at eta tends to Zero F dash is zero and theta tends to zero okay,
so these are the equations and boundary conditions for natural convection from uniform wall temperature,
okay. Next let us show you that what happens in
case of large Prandtl number case and small Prandtl number case okay, in case of large
prandtl number case the previous equations will be changing in this form, F triple dash
plus theta equals to zero plus theta triple dash plus three F theta dash equals to zero
okay, and for small prandtl number these will be becoming three F F double dash minus two
F dash square plus theta equals to zero and theta double dash plus three F theta dash
equals to zero corresponding boundary conditions I have written over here which are more or
less same okay, at eta tends to zero, eta equals to zero F and F dash both are zero,
theta equals to one, eta tends to infinity F dash and theta equals to zero and for prandtl
number tends to zero. At eta equals to zero F is zero and theta
equals to one and as eta tends to infinity, F dash and theta both tends to zero corresponding
Nusselt number also I have also written if we solve this two equations numerically then
we can find out Nusselt number becomes point five zero three Rayleigh number to power one
by four okay, so this is rally number is very important this is nothing but Grashof into
prandtl number, okay. And here in this case prandtl number tends to zero we give Nusselt
number equals to point six Rayleigh nuumber to the power one by four Prandtl number to
the power one by four, okay. In summary let us see what we have learnt
we have seen what is the choice of velocity scale for non- dimensionalization, we have
seen u not is nothing but root over of g beta L delta t okay, so delta Tw minus T infinity
as wall temperature was known. Governing equation for natural convection around vertical hot
plate at constant temperature so this was the generalized governing equation, this is
from the momentum equation, this is from the energy equation and corresponding boundary
conditions this we have derived. Different Nusselt number correlations also
we have mentioned for low prandtl number and high prandtl number case, in low prandtl number
case point six was the co-efficient and for large prandtl number case it was point five
zero three. And in case of large prandtl number Rayleigh number came into picture, okay here
also Rayleigh number came with Pr as having power of one by four, okay. Now let us test
your understanding what you have understood in this lecture. Help me in identifying the velocity and temperature
profiles for natural convection in different prandtl number limits, okay. I have shown
here two cases so red lines are for temperature profile and black line is for the velocity
profiles, okay. So here temperature profile very promptly goes to zero okay, but velocity
profile prevails and it is having a very high velocity profile so delta is high compared
to delta t. And in this case we are having delta t higher
compared to delta okay, so you need to tell me which one is for large prandtl number,
which one is for small prandtl number so I think all of you can guess the correct answer
is first this one is for large prandtl number okay, so that is why velocity boundary layer
is having higher thickness compared to thermal boundary layer just opposite happens over
here in case of low prandtl number, okay. So with this I will end this lecture, thank
you in my next lecture we will be having natural convection from a plate which is having uniform
hit flux in between if you are having any query please keep on posting in the discussion
forum, thank you.

## One Reply to “Natural Convection: Uniform Wall Temperature”

1. Sunal Gautam says:

Aata Majhi Satakli