Video summary
The lecture focuses on the mathematical derivation and non-dimensionalization of the governing equations for natural convection, building upon previous discussions about the Boussinesq approximation. By introducing a reference velocity, $U_{\infty}$, which does not physically exist in pure natural convection but serves as a scaling parameter, the instructor demonstrates how to compare the relative magnitudes of buoyancy and inertial forces. This comparison leads to the definition of the Richardson number ($Gr/Re^2$), which acts as a critical indicator for determining the flow regime: when this ratio is much greater than one, natural convection dominates; when it is much less than one, forced convection prevails; and when it is of order one, mixed convection occurs. The analysis also highlights that while the energy equation simplifies similarly to forced convection by neglecting certain terms, the coupling between temperature and velocity fields remains essential for solving the coupled continuity, momentum, and energy equations simultaneously.
Following the theoretical derivation, the discussion shifts to practical correlations used to calculate the Nusselt number for various geometries, such as vertical plates, horizontal surfaces, and cylinders. A key concept introduced is the film temperature, defined as the average of the surface and ambient temperatures, at which fluid properties like viscosity and thermal conductivity should be evaluated. The lecture emphasizes that empirical correlations often take the form of the Rayleigh number ($Ra = Gr \cdot Pr$) raised to a specific power, typically $1/4$ for laminar flow on vertical plates or $1/3$ for turbulent regimes. Special attention is given to horizontal surfaces, where the orientation of the hot surface significantly impacts heat transfer efficiency; specifically, a hot surface facing downward results in lower heat transfer coefficients because rising warm air gets trapped beneath the plate, whereas a hot surface facing upward allows free convection currents to develop more easily.
The final portion of the module applies these principles to solve real-world engineering problems, such as calculating heat loss from hot water pipes and square plates in rooms. Through detailed examples, the instructor illustrates how to compute the Rayleigh number based on the characteristic dimension (diameter for cylinders, height or area-perimeter ratio for plates) and then determine the convective heat transfer coefficient. A crucial takeaway from these problems is that natural convection is almost always accompanied by radiative heat transfer; neglecting radiation can lead to significant errors in estimating total heat loss, as demonstrated by a case where radiative losses were nearly equal to convective losses. The lecture concludes by comparing different scenarios for a single plate, showing that the orientation of the hot surface drastically changes the heat transfer rate, with the lowest efficiency occurring when the hot side faces downward due to the stagnation of warm air.
Read the full video transcript
In the last module, we looked at the
derivation of the governing equation and
we came to the point where we used the
booziness approximation and wrote the y
momentum equation after Simplifying the
d² v by d y² term due to scaling using
the booziness approximation we got the
buoyancy force terms in terms of the da
and the delta t terms and the energy
equation also got simplified because d²
t by d y² was neglected. So this
equation now unlike your force
convection this coupling of temperature
is directly seen here. Okay. And these
are the boundary conditions. Let's go
further. Uh we'll come back to this
definitions. So we have a scaling
analysis or we have the
non-dimensionalization which we need to
do and I will go through this and uh
afterward there are a couple of
correlations that need to be shown to
you. So what are the what is the
non-dimensionalization?
If we take the uh length of the plate
the vertical direction is h just as we
had for the external flow l was the
length of the plate. We are calling this
as H. We could call this as L also. Let
us define.
Let us define XAR
is equal to X by H. Y* = Y by H. Length
scales are there. T* is equal to T minus
T infinity divided by T S minus T
infinity.
uh
u
I don't want to make a mess. U* is equal
to U by U infinity. V star is V by V
infinity. So U* is equal to U by U
infinity. Var is equal to V by U
infinity.
Okay. Now one might ask a very simple
question. What is u u infinity here?
Because we said natural convection. This
is the plate. This is my velocity. This
is my gravity. This is my t infinity.
What is this u infinity? This u infinity
is essentially introduced here so that
we understand what is the
non-dimensional number that is going to
come and where force convection and
natural convection are important. So the
relative magnitudes of the force
convection component and the natural
convection component are going to be
obtained by use of this so-called u
infinity. Okay. So we'll we'll go
through the maths now. Continuity
equation is du by dx plus dv by dy =0.
X momentum equation or sorry Y momentum
equation is U DV by DX plus V DV by DY
is equal to mu D sorry mu D²
V by DX²
+ row minus
uh sorry
row G tus T infinity and energy equation
U DT by DX plus V DT dy dy is equal to
alpha d² t by dx². So these are my
governing equations. So let us
substitute whatever we have from the
non-dimensionalization.
So du by dx + dv by dy becomes h into
u is
u infinity. So I'll just write this
here. uh u infinity du*
divided by h into dx*
plus u infinity dv*
divided by h into dy*
equal to zero which means du* by dx*
plus dv* by dystar =0 just like before.
Okay, let's go and do the other one.
This is u infinity into u star
u infinity into dv star divided by h dx*
plus u infinity v star u infinity dv*
divided by h into dyar
is equal to new /
H²
into
DX*²
U infinity
Var
U infinity V star D Var
uh sorry U infinity DVAR
divided by DX xar²
d² v star. Okay, remember this d by dx
of dv by dx. So this would be d by dx of
u infinity dv* divided by h dxar.
Second time this would be replaced by h
into dxar. That is why you get h². Okay.
So this one and right hand side is row
into g t minus t infinity is tar into t
ss minus t infinity. So I get here u
infinity squar by h u* dv* by dx*
plus
v* dv* by dy*
is equal to new
u infinity by h²
d² var by dx*
squ plus row g sus t infinity t star.
Now you divide through by this etc. I
would I'm just writing this part as it
is. Divide
by u infinity squar by h. So you lose
that here. So u infinity squar by h you
would get new
h
u infinity
b ² v star by dx*
squared
plus row g t s - t infinity into h by u
infinity²
into t star.
Okay.
Okay. Let's just go and check whether we
are doing everything right. Yes, we are
doing things right. And what I'm going
to do next is this is your Renaol's
number. I didn't write this here. This
is just going to be U² DVAR by DXAR
plus Var DVAR by DY*
equal to this one. This one is 1 / r e h
correct? R e h is u infinity h / new.
This is 1 / reg. So 1 / e h d² v star by
dx* squ exactly like what we got in
force convection plus here is the
important part row g
t s minus t infinity what do I need
row g you had uh
where I think row g this one
I go back here I think row is row was I
think I made a mistake here. It's G beta
sorry. So uh
G beta delta T sorry. So this is we had
substituted G beta delta T right. So
this should be where is the equation? G
beta delta t not delta t. G beta delta
t. So g beta delta t. Just make a change
here. Beta beta.
Okay. So this is G beta delta t
divided by u infinity squared. So I have
u infinity squared. I would want uh
multiply and divide by
h².
So this will be h cubed u infinity squar
h²
multiply and divide by new ² and t star
remains as it is. What did I do? I want
this to be like a reol's number.
This one to be like a reol's number. I
had a h so I multiplied and divided by
h². So h cq I multiplied and divided by
new ^ 2 so that this become squ and this
group of terms
g beta deltat t h cubed by new ^ 2 is
also a non-dimensional quantity and this
also is your reol number. So u*ar
dv* by dx*
plus var dv* by dy*
is equal to 1 by h d² var by dx* squ +
g beta delta t h cubed by new 2
divided by
u infinity h by new the squared t star.
This group of terms is called as the
grash of number
GR of H which is G beta T S minus T
infinity
characteristic dimension cubed by U² and
of course this is your RH
so this is H² squ. So this is nothing
but 1x h d² v star divided by dx* squ +
g r of h over e h²
into d star.
This quantity plays a very important
role in understanding whether your
buoyancy force is important or not. And
that is why we had to have a U infinity.
U infinity doesn't feature here at all
but it features in the Renaol's number
definition. And if GR by R H²
is much greater than one which means the
buoyancy force term is dominant. So
natural convection is dominant because
of this one.
If GR by E² is very less than 1 that
means force convection is dominant.
This is what happens when you have force
convection U infinity is large is very
large. So even though natural convection
is there all the time natural convection
is there this ratio becomes very small.
So we can neglect the buoyancy force
term. Third one of course GR by E² is
order one mixed convection that is
natural and force convection both of
them exist that is why this group of
term is very very important called as
Richardson number
okay so that's what is told here in this
set of slides now if I go through the
energy equation in a similar manner I'm
not going to walk through this you get
exactly what you got in the force
convection part it is PR r 1 / P R as
you can see here 1 / P R exactly that in
fact this term would go away this term
would go away and you'd have only this
term left okay I'm not walking through
this everything is there in the slide so
what is the consequence of all this you
have continuity you have x moment you
have y momentum you have energy equation
which on non-dimensionalization becomes
these three equation 1 by this 1 + G r².
If this term was neglected, if this term
was neglected, which means buoyancy
force is very small, we get only the
force convection equation which we know
already. Right? So whatever you see here
the additional term for natural
convection is this one. That is the only
extra term that you are getting. And how
are we going to solve this? Mass,
momentum and energy equations are
coupled. As you can see, temperature is
right here. You solve for u v and t from
which you get your delta delta t nasalt
number and non-dimensional velocities
non-dimensional temperatures are all
functions of reol's number pralle number
and rally number what is rally number
rally number is nothing but gr into pr
this set of slides gr into pr is your
rally number this multiplied by new over
alpha would give you alpha new that's
the only difference. So GR R into P R is
your RL number. These non-dimensional
number definitions you'll have to
remember. Reynolds we know Pantle we
know G beta delta TL cub characteristic
length cub divided by new ^ 2 is your
grash grash into pantle is your rally
number. Okay. And this is what I had
already told uh the role of GR by squar
force convection where it dominates etc.
For vertical plates critical grashoff
number is about 10 ^ 9 after which the
flow becomes turbulent.
Okay. So with this we are done with the
derivation. Now just how to solve these
problems. Nassus is typically a function
of we got we got this from our uh
non-dimensionalization.
Nassus number was a function of PR in
case of force convection. Now you have
this additional quantity which is
P prash
of number. GRPR is rally number
would vanish in natural convection
alone. So natural number is only a
function of GRPR or rally number. And
your correlations typically would be of
the form C ra to the power n or m and
almost always you would see ra to the^
14. This can be shown by scaling
analysis by the way. Okay not going into
the details of it. So gr is this ra is
this and typically the properties of the
fluid are evaluated at what we call as
film temperature which is t surface plus
t infinity by 2.
Now we look at the correlations for
various geometrical situations. Vertical
plate which is heated uh typically RA ^
1/4 for this range of 10 ^ 4 to 10 ^ 9 9
to 13 RA ^ 1/3 and this big equation 6.3
is generally used because it's more
complicated to use but reason much more
accurate. And if you have an incin
plate, you replace g by g cos theta.
Okay, very simple. Now whatever we are
doing are all empirical. If I have a
plate which is hot on top, okay, and
cold at the bottom, the hot convection
would be happening this way. Okay, hot
air will rise up etc. That is the first
diagram. If I have hot surface facing
downward and top surface is cold, the
convective currents will all be at the
bottom. For it to rise up is going to be
difficult because they have to come
around the surface. Okay. So, it's going
to be a slightly different convoluted
way of doing it. The correlations for
Nusel number will be different. You can
see here for the horizontal plate
surface area A and perimeter P upper
surface is hot. The characteristic
length in both cases is surface area by
perimeter. So if it is a circular plate
p r² divided by 2 pi r square plate
appropriately both are same the nusle
number if you see 54 e to ^ 1/4 and here
it is 27 roughly half let us say I am at
10 ^ 6 rally number and you have 0 54
and 27 the nusled number for bottom
surface being hot is lower And that's
natural, right? The convective currents
are not able to go out freely.
Therefore, the heat transfer coefficient
is going to be lower. Okay?
Flow around a cylinder. And you have
flow around a cylinder which is this
way. You have the nasal number given
here. Uh this is flow around a sphere.
This is cross flow around a cylinder.
Vertical cylinder can be treated as a
vertical plate when this condition is
satisfied. So most of these natural
convection problems will involve
choosing the right coefficient uh sorry
right correlation
calculating rally number putting it in
the appropriate formula for nasal number
H LC by K get your H get your Q which is
H into delta T into A that's it two
problems are there I'll go through one
and if uh we can leave the other one for
you to do a 6 m long section of 8 cm m
diameter hot water pipe shown in the
figure passes through a large room whose
temperature is maintained at 20°. So I
have a pipe. Many times you would have
seen you have hot water pipes in you
know large uh toilets, basement of
buildings etc. They'll carry hot water
and the surrounding room is at 20°. The
pipe surface temperature is 70. So there
will be continuous loss of energy. You
want to calculate that. So in your
conduction problem etc. This would have
been given as a convective coefficient
10 or something and you would do this
problem. Right? Now you have to
calculate that quantity. That's it. So
two temperature are given T surface and
T fil uh T infinity. So calculate the
film temperature which is 45°
centigrade. Properties are evaluated at
this beta is nothing but 1 / film
temperature. Please remember this is
approximate as 1 / film temperature. How
do I get it? You go back to the
definition of beta and you will you'll
be able to get this. Okay. So delta row
by row into delta t that delta row by
row is approximately constant and this
is approximately one and this is one
over film temperature. Okay. So all
these properties are known. First step
is calculate your rally number. What is
this? This is a pipe which is suspended
in the room. So it's cross flow over a
cylinder. The characteristic dimension
is your diameter of the cylinder. So
rally number g beta delta t dq by alpha
new delta t is t s minus t infinity and
substitute all the values here you get
your rally number as 1.869 869 into 10 ^
6. Substitute into the appropriate
correlation for Nassel number which is
given here this one 6.7
and do the maths lot of mistakes are
possible especially in the denominator
term do it step by step. So this whole
thing to the power 9x6 and the whole
bracket to the power 8x 27 somehow there
will be a tendency to make this you know
uh 1x 6 that's not because there is a
one here and this is only to this power
so be careful mistakes happen only in
this step so nasal number is 17 and you
put this for your hlc by k
characteristic dimension is d so hd by k
is 17.4 4 from this you'll get h is a
mere 5.87 87. Look at how small these
numbers are. More than the actual
numerical uh solution of the problem.
You need to look at these h values.
Natural convection of air. The in the
first SL set of videos that we generated
we had a table of H and gases natural
convection H was given to be 2 to20 or
something. Okay. We are having a H value
of 5.8 8 which is very very small and
therefore the heat transfer is poor. So
Q is H A deltat T. Substitute the number
443 W of energy is being lost by the air
uh by the water through natural
convection. Now if you include a
radiation here will become important.
Yeah that's there in the next part. Pipe
will lose heat by radiation as well. So
sigma epsilon a t surface minus t
surface to the 4 minus t infinity ^ 4.
Let's say emissivity is 1. You have 553
watt which is being lost 443 by natural
convection. Equal emissivity even if it
is.8 you would have 400 450 W. So
roughly equal quantity. If I neglect
radiation I am losing out on
information. Actually the pipe is losing
heat at the rate of about 900 W but only
natural convection I will say 450 okay
450 W perh what is being lost so I will
insulate it appropriately but still I
would be losing heat because of
radiation and that would not have been
accounted so the important thing from
this simple problem is natural
convection is always almost always
accompanied by radiation both and it's
true the other way also So emity of real
surface is less than one radiation heat
transfer will be slightly lower than
this but it'll be significant as most
systems cooled by natural convection.
Therefore radiation analysis should
normally accompany natural convection
analysis unless the emissivity is really
low. Okay.1 or something you can neglect
the radiation part. Okay one more
problem. So you have a square plate
which is thin 6 m by 6.6 6 m placed in a
room at 30°. One side of the plate is
maintained at 90, the other side is
insulated. So this is a classic case of
top plate, top surface hot, bottom cool,
and vice versa, which is what this
problem is. Determine the rate of heat
transfer from the plate by natural
convection. If it's made vertical, if it
is the horizontal surface with the hot
surface facing up or the hot surface
facing down. Three cases, same numbers.
So steady state air is an ideal gas at
low local pressure is one atmosphere.
Properties are evaluated at film
temperature which is average of 90 and
30 which is 60°. Go to the table. Beta
is 1 / film temperature 1 / 333 K
prantle everything is known. Once you
get this these are the three cases. So
vertical plate is the classic case where
you have a boundary layer which is
formed and that we know how to
calculate. The characteristic length is
H. For these cases B and C the
characteristic length is AC by P. So sub
calculate your rally number G beta delta
T characteristic length Q by alpha new
and three cases three different
characteristic length everything else is
the same. So this is a vertical plate
it's 6. Uh in the other case it is a S
by P which is L² by 4 L. perimeter is 4
L² plate surface area is L² LX4
that's what is here also so r number is
going to be calculated from this G beta
delta T h cubed by alpha new you get uh
there is
which is going to come as 7.65 65 into
10 ^ 8. So you go to the case with
vertical plate. I'll go back to the
correlation slides. You have vertical
plate correlation. This big correlation
6.3 is being used and I will substitute
the numbers here and get nasalt number
is 113.4.
Lot of scope for mistakes in terms of
calculation nothing else. So procedure
is simple. there renolantal
nasult h here rale nasult h that's it
so nasel number is calculated even by
the simple form if you're hardpressed
for time okay you have 5 minutes and you
want to do the problem this 0.59 ra to
the^ 1/4 will also do the job what is
the difference 113 versus 98 okay 15%
yeah 13% lower doesn't matter you'll not
get zero but it's okay. So nasal number
is calculated h is five again last
problem it was five here also it is five
5.8 there 5.3 here area is l² so q dot
is 115 watt go to the other case lx4 is
the characteristic length calculate r
number 1.19 into 10 ^ 7 here it is 7.65
65 into 10 ^ 8
you have NASL is.59 RA ^ 4 31 here 98
and 113 so much smaller nasal number
much smaller of course the
characteristic dimension is changing
therefore your H is slightly larger 5.9
and Q is 128.4
Q was here 115
128.4 4
and last case it is about half so 64.27
2.27 27 RA ^ 4. This was 59 RA ^ 4.
Okay. So roughly half and it gives me a
value 273
H value. So natural convection heat
transfer is lowest in the case of hot
surface facing down. So the this is not
surprising because hot air gets trapped
under the plate and cannot get away. It
has to go around. So if this is the
plate, it has to go around like this for
convection to be established. Whereas if
the top is hot, it will just go and mix
like this. So that is why H is lower for
that case. Okay. So with this we come to
the end of the module on natural
convection.
Then we have of course two very
important topics. One is heat exchanges
and the major chunk is on radiative heat
transfer. Thank you.