Video summary
The lecture begins by contrasting two fundamental scenarios in internal flow heat transfer: constant wall heat flux and constant wall temperature. In the case of a constant wall heat flux, the bulk mean temperature increases linearly along the pipe length because the heat transfer rate per unit area remains unchanged. Consequently, the slope of the wall temperature line matches that of the bulk mean temperature line in the fully developed region, resulting in parallel lines. However, in the developing region where the heat transfer coefficient decreases with distance, the temperature difference between the wall and the fluid increases, causing the lines to diverge before merging into parallel lines once fully developed flow is achieved.
The analysis shifts to the constant wall temperature case, which presents a different physical constraint where the surface temperature is fixed rather than the heat flux. Under these conditions, the energy balance reveals that the local temperature difference between the wall and the bulk fluid decays exponentially along the flow direction. This exponential decay is governed by the Number of Transfer Units (NTU), defined as the product of the heat transfer coefficient and surface area divided by the mass flow rate times specific heat capacity. The lecture emphasizes that while increasing the tube length continues to raise the fluid temperature, the marginal gain diminishes rapidly; typically, an NTU value around five is considered sufficient for engineering purposes because extending the pipe further yields negligible temperature improvement at a high economic cost.
To quantify heat transfer in this scenario, the concept of Log Mean Temperature Difference (LMTD) is introduced as a critical tool for heat exchanger design. The LMTD accounts for the varying temperature difference between the fluid and the wall along the length of the pipe, proving to be significantly different from a simple arithmetic average. The total heat transfer rate is expressed as the product of the overall heat transfer coefficient, surface area, and the LMTD, linking this internal flow analysis back to broader thermal resistance concepts like those seen in conduction problems.
The session concludes with a practical example involving water being heated by condensing steam inside a copper tube. By calculating the required heat load based on the water's mass flow rate and specific heat capacity, and determining the LMTD from the inlet and outlet temperature differences, the necessary surface area is found to be approximately 4 square meters. Given the tube diameter, this translates to a pipe length of about 37 meters, illustrating how long tubes are often required for significant temperature changes under constant wall temperature conditions. The lecture ends by previewing the next module, which will derive expressions for the Nusselt number for both constant heat flux and constant wall temperature cases using fluid mechanics principles.
Read the full video transcript
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In the last module, we did the general
analysis for energy balance. We started
with a control volume. Uh wrote a series
of assumptions. Use the general energy
equation. E dot in minus E dot out plus
E dot generated plus E dot store is
equal to E dot stored. Did the analysis
and said for a constant wall heat flux
case DT by DX remains constant and
therefore the line of bulk mean
temperature is a straight line. The
variation of bulk mean temperature is a
straight line for constant wall heat
flux case. That is not the case when you
have varying heat flux. Okay. Then we
also went on to say that qp prime is h
deltat t in in fully developed case h is
independent of x. So this is a constant
and therefore ts minus tm is going to be
constant which means the slope of the
wall temperature line would be the same
as the slope of the bulk mean
temperature line in the fully developed
region. Therefore the lines will be
parallel in the developing region. We
also we realize that h is going to vary
this way just as friction factor varied
and h is going to decrease which means
in this part qp prime by h it's a
decreasing function. Therefore t ss
minus tm is an increasing function in
the developing region. Therefore we got
a diverging set of lines in this part.
These two will merge at this location of
fully developed flow and then they will
remain parallel. Okay. We also just
looked at setting up the problem for
non-uniform wall heat flux case and also
if you had a situation where the dimen
dimensions or diameter of the channel
was also changing. Let's just take a
look at the slides and see what the
slides have for us. So this is the
derivation and we just looked at it this
part and we said that
qp prime is uh dts by dx is dtm by dx.
We said the lines have same slope and
from that what did we say? We also had
the condition for thermally fully
developed. This is the condition for
thermally fully developed. So if I write
this in this form we get local
temperature gradient is equal to wall
temperature gradient. And we also said
wall temperature gradient is same as
bulk mean temperature gradient. dtm by
dx is equal to dts by dx which is equal
to constant. Okay. So now we have all
the three gradients dts by dx dt by dx
dt m by dx are all constants and that is
nothing but equal to qp prime into pi d
by m dot cp perimeter is pi d for a
circular pipe. What is important in wall
temperature is I mean sorry wall heat
flux constant the gradient the slope of
these curves are identical. So you will
see exactly the same shape. That means
the gradients will be identical. The
thickness of the profile will change
depending on whether there is heat added
or heat removed. So fully developed flow
in a tube subject to constant surface
heat flux. Temperature gradient is
independent of X that the shape of the
temperature profile does not change
along the length of the tube. Important
thing from fill in the blanks concept
questions point of view.
Okay, this two profiles are shown and
the profile the gradient the slope is
nothing but this. Okay,
and of course we wrote this uh parimeter
is pi d mass flow rate is row into a
into bulk mean velocity a is by 4
d². So when you put all this you get
this as 4 into qp prime divided by row u
m cpd which is equal to constant.
Okay. Next we go to constant wall
temperature or surface temperature case.
This is a interesting case because this
you have a constraint on the wall
temperature. So nothing comes free in
life. Okay. So either you have the wall
heat flux constant or the wall
temperature constant. So what do we do
here for this analysis? So let's quickly
go back to our set of slides.
T- wall equal to
constant case.
Let us draw the temperature profile
here. This is going to be used in your
heat exchanger chapter which will be
taught later. So whatever we are doing
now internal flow the direct application
of this is heat exchanges. Okay
temperature length if T wall greater
than T bulk mean then Q is added into
the fluid.
So the line of T wall will be higher
than the bulk mean temperature. You'll
have this somewhere here. If T wall less
than T bulk mean Q is removed from fluid
right
okay so for constant wall temperature
case what is the energy balance exactly
the same E dot in minus E dot out plus E
dot generated is equal to E dot stored.
Again we are talking of no heat
generation steady state. So MCP
Tbulk mean
the same diagram as what we drew for the
heat flux constant. I'm not drawing I'm
not going back and forth because we want
to be on the same slide. So this is your
control volume of thickness dx. Instead
of qp prime this is t wall is equal to
constant or t surface equal to constant.
I think we using t surface. So let's
just go back and use t surface here. t
surface is equal to constant.
Okay.
Same thing. M dot CP bulk TM DTM
TM
+ DTM.
So e dot in this one
plus
qp prime pi dx
is equal to m dot cp tm + ttm
which now will give me m dot cp dtm is
equal to qprime
pi dx.
Therefore, dt m by dx as before till
here nothing has been told about the
boundary condition I am just writing
energy balance. Okay. Now from here
onwards things will differ. Earlier we
said if t uh if wall heat flux was
constant the entire term is constant and
you'll get something. Now we are saying
this
is not constant
not known but what is known qp prime is
equal to h into t ss minus tb bulk mean
of x.
Okay,
fully developed case
h is equal to constant.
Okay, so
this we are saying is constant force to
be constant
imposed.
Right? So I will write this as dt m by
dx
is equal to h into t surface minus t
bulk mean of x into pi dided by m dot cp
which I'm going to separate the
variables dtm divided by t s minus tm
of x is equal equal to h pi d by m dot
cp. Now I will do some circles here.
ts is constant. So let
t s minus t of x be = theta. dt m is
equal to d theta
at x = 0
t = t m i theta is = thetai i at x = any
x t = tm of x theta is = theta sub x so
these are my transformations so I will
put d theta pi theta the minus sign will
take it to the other side minus h by d
by m dot cp implies log theta
theta i uh sorry this is
uh pi dx
was missed
okay theta is equal to minus H pi D by M
dot CP this is you have X
H is not a function of X we have assumed
here
okay so that has been inherently assumed
so this would transform to log of theta
by theta I is equal to - h pi dx by m
dot cp which is - h into perimeter x by
m dot cp. This means theta is equal to
theta by theta i =
e xp
minus hpx by m dot cp that is theta is
equal to t s minus t bulk mean local is
equal to t s minus t bulk mean at inlet
maximum
time exp
minus hp PX by M. CP exponential decay.
Where else have we seen the exponential
decay? Transient conduction. Okay, same
concept only here also the exponential
decay is there. And therefore the
local temperature difference this is
theta of X. This is theta at
I.
this one
theta or temperature bulk mean
temperature
this is what it is so dtm I've
substituted all that and this is going
from 0 to l so the integration is from 0
to l we did from 0 to general x and we
will see the temperature difference
between the fluid and the surface that
is theta decays exponentially along the
flow direction
and it depends on this magnitude H into
P into X / M dot CP P into X is nothing
but the surface area so we have surface
area dependency directly coming in and
if I translate this into the
temperatures I get local temperature is
TS minus minus maximum temperature
difference e ^ min - hpx by m cp and
this quantity we just write this in a
form we like so t
therefore from here I will say ts minus
tx is equal to ts minus t i exp
minus hpx by m dot
CP
which would give me T bulk mean of X is
T S minus TS - T M I E XP - HP X by M
dot CP
local temperature
bulk mean temperature.
Okay. Now TM at L or the exit
T S minus T S - T M I E XP
minus H into P L P P P P P P P P P P P P
P P P P P P P P P into L is the total
surface area of the circular pipe.
Perimeter is pi D. L is the length. PDL
which is the surface area for heat
transfer. So this is nothing but
heat transfer area.
Okay. So
minus H A by M. CP this controls
controls the temperature distribution.
This one HAS by M.CP CP has a huge
impact or has a very very very high
significance in heat exchangers as well
as in internal flows and we will see
what that means very quickly and it's
important also
very important h by mp is a
dimensionless quantity anything inside
the exponent has to be dimensionless we
call this as ntu or number of transfer
units
It is a measure of how effective the
heat exchange is there or heat transfer
system is operating. So, NTU of about
five indicates the limit has been
reached for heat transfer. We have done
a simple problem here. Tb bulk mean is
20° and T surface is 100. You're seeing
what is the exit temperature. See when
you have this kind of a
problem where the temperature as I go
further and further in length it will
asytote as L tends to infinity T bulk
mean as totes to TS
right so that means each additional
incremental delta X that I am adding
that will have a very very small impact
impact on the
temperature of the bulk fluid. So it's
law of diminishing returns kind of
stuff. So when you when you initially
when you added a small delta x it caused
a significant effect in the bulk mean
temperature. But as I go on adding
length further and further, a stage is
reached when the change in temperature
is so small compared to that dx added.
That means its utility goes on
decreasing as I add more and more delta
x thicknesses. And that's what is
exemplified by this statement when we
say when you do a calculation of this.
So theta is by theta I is equal to E XP
minus H A by M dot CP. So that hs by m
cp you're going to just change and theta
by theta i theta is uh theta i is 100 -
20 which is 80 t e is 100 minus whatever
that te which we are calculating is the
theta you will write this ntu is just
being changed so exp
let me just show it instead of uh this
one here so theta Theta local or theta
exit divided by 80 is equal to exp minus
h a s by m dot cp this quantity whole
bracket I am changing
from 0.01 01 to 0.11
2 3 4 5 based on this I calculate theta
at exit
which is T surface minus T bulk mean at
exit from this I'll get T bulk mean at
exit is T surface
minus
theta I into this particular quantity
Right. So what's happening is that when
I go on increasing this NTU, H A S by
M.CP, H is typically constant. M dot CP
are all constants. The only way HS by
M.CP increases is by increasing the area
which means by increasing the length. So
for NTU of.5
exit temperature is 51.5 NT of 1 is 75
it's 99.5
NTU of 10 is giving you exit temperature
100 but please understand this five unit
or 5 m increase in length for a 0.5°
centigrade rise in temperature. Is it
even worth it? Because when you're
having fluid flow, there will be
friction loss. For first 5 m length, you
got 99.5°
centiggrade as a temperature. Next 5 m
length, you're getting an additional
0.5. You would be senseless to accept
this. So typically about five NTU of
about five, we get 99.5. And that's okay
from engineering point of view because
it's the cost involved, the space
involved, the weight involved in going
from NTU of 5 to NTU of 10 for a
small.5° rise in temperature is not at
all worth it. Okay, that's what this
problem is trying to say. A large NTU
and thus a large heat transfer surface
area which means a long large tube may
be desirable from a heat transfer point
of view but may be unacceptable from
economic point of view that means cost
space everything would be a problem.
Okay. Okay. So then this is the nature
of the graph is shown here. The
temperature distribution logarithmic
one. So h by m dot cp. So m dotcp is hs
by this delta t. This is just a
manipulation. So q dot is m dot cp into
t exit minus t inlet. T exit minus t
inlet will retain as it is. M dot cp we
will substitute from here to here. So
minus h a s into t minus ti divided by
some particular term. And this TI minus
TE this minus sign gets absorbed here.
TI minus TE. I add and subtract TS from
this. So TS minus TS. I group them as TS
minus TE and TS minus TI. Okay. So Q is
equal to H into A as S into this
particular quantity which is just a
manipulation of the numerator and this
particular quantity
exit temperature difference minus inlet
temperature difference divided by log of
exit minus exit by inlet temperature
difference. This temperature difference
this whole thing is called LMTD or log
mean temperature difference. Log mean
temperature difference is essentially
this minus this ts minus ti minus ts
minus t divide by log of ts minus ti /
ts minus t. So whichever term you have
here first that is going to be there in
the numerator of the log term. This is
very very important. This derivation is
again a part of your heat exchanger
design. You don't have to do it again
separately. We have done it here
already. For constant wall heat flux
case, we are getting this kind of a
thing. You can extend this not to
constant wall but approximately constant
wall case also. Okay. Where the
temperature variation of T- wall is
negligibly small. Log mean temperature
difference is very important definition.
So, LMTD
or log mean temperature difference
deltat t log mean is nothing but deltat
t in minus deltat t out divided by log
of deltat t in over deltat t out vice
versa also.
So this is my diagram of interest. This
is in deltat t in. This is my deltat t
out.
Okay. Whichever you take first that will
appear in the numerator here.
And this is log mean temperature
difference representative delta t for
the entire pipe of interest.
Okay.
And we'll just do this problem.
This
we also said h a delta t.
Please remember this has been done for
inlet to exit. That's why everywhere t
is there as exit temperature. You could
do it from inlet to another arbitrary
location x where the temperature is t
bulk mean at that x location. Okay. So
that is left as an exercise for you to
do. But one more very important thing
which I wanted to share is this one H A
delta T log mean right so this is Q
which can be delta T log minide 1 / H A
divide delta T log minide 1 / H A
[snorts] uh and this 1 / H A is like
your thermal resistance remember when we
did our conduction heat transfer
deltat t over
everything. I will just write this
is Q and this everything was my thermal
resistance.
And we also said Q in general is del U
into delta T. UA into delta T where we
said delta T is
and therefore U A is 1 / summation of R
thermal. This was something which we saw
earlier in conduction also. Okay.
Parallel resistance, series resistance
etc. Now what are we having? Q is equal
to H A S deltat T log mean some delta T.
This is can also be extended as U A
delta T log mean which should be deltat
T log mean divide by 1 / UA.
Okay. So we link what we had studied in
conduction for resistance in series
parallel to a concept similar here. Log
mean temperature difference and this 1 /
UA is nothing but reciprocal of thermal
resistances again. Okay. So this part
just wanted to reiterate here.
Let's look at the simple problem. Water
at uh enters a 3.5 cm diameter copper
tube thin copper tube which means wall
thickness is negligible at a rate.3 kg
per second and is heated by condensing
steam at a temperature of 110. Steam
condenses at 110 on the outside which
means the wall temperature is 110. If
the average heat transfer coefficient is
900 W per meter² Kelvin, determine the
length of the tube required to heat the
water to 105° centigrade. So you have
inlet at 20, outlet at 105, wall is at
110. Okay. So we want to calculate the
length of the tube steady state constant
properties. Heat transfer coefficient
given is known and constant. Conduction
resistance in copper tube is negligible.
So that the inner surface temperature is
equal to the condens condensation
temperature of the steam which means you
do not have thermal resistance between
the outer and the inner wall which means
it is a thin wall tube which is what is
stated and therefore you don't have to
worry about that part but that is going
to affect your temperature. So the inner
surface temperature is nothing but the
condensation temperature of the steam
which is 110. So it's as if you're
having a problem as if you're having a
problem with T wall is 110.
You want this to be 105 at exit 20°
centigrade fluid is coming in. What is
the length required? This is your
temperature profile. This is the X. What
is the length required to achieve this?
That's what is the problem. Very simple.
Okay. So, specific heat
of water is taken at a bulk temperature
20 + 110 by 2 which is 65 and it's 4187.
Heat of condensation of steam at 110°
centigrade has been found from your
steam tables as 2230 kiloj per kg. So to
achieve this increase in temperature
from 20° to 105 what is the length if
this is the condition? So heat added to
the water is m dot cp deltat t of water.
So m dot is.3 cp 4187
deltat t is 105 minus 20 the answer is
106.77
kilowatt. So 106.77
kilowatt is added to the water. How is
it done? By constant wall temperature
case steam is condensing that condensing
steam is giving heat to this. So we want
to find out now lmtd. So we draw draw
the sketch of the temperature
distribution 20 105 and 110. Deltat ti
is 110 minus 20 which is 90. Delta TE is
110 minus 105 which is 5°. Delta T log
mean is nothing but the difference of
these two divide by log of this one. So
whichever you have taken first that is
what will appear in the numerator. Based
on this the log mean temperature
difference is 29.41.
Look at this 110us 20 90° is the
temperature difference at the inlet. 5°
the temperature difference at the
outlet. It is not It is not 80 + 5 by 2.
No, it is not the arithmetic average. So
you cannot say this is 80° here and this
is 5°. 80 + 5 / 2 42.5. No, the log mean
temperature difference is 29.41.
Far away from 42.5.
Having known this, use your Q is equal
to H A delta T. H A delta T. Everything
is known to you. So you get the surface
area as 4.03. 03 m because you want to
calculate the length. Now surface area
is p into d into l. So you know the
diameter from that you can calculate the
length. 37 m of piping is needed for you
to achieve this increase in temperature
by 85° centigrade. So water really
imagine it's a very very very hard to
eat water. So 37 m of pipe length is
needed. Simple problem where it's a
direct substitution of the values into
the problem. Important thing to note is
the use of deltat t log mean and it is
not deltat t inlet plus deltat t outlet
by 2. It is not the case. It is a log
mean temperature difference. you use it
ti minus t or t minus ti whatever you
use the value of delta t log mean will
come out to be the same in this problem
you could be given a length and asked to
find the exit temperature of the water
exactly the same way except that now
instead of finding the log mean
temperature difference you would have
the
uh lmtd calculated from
after obtaining the uh exit temperature
you cannot do it that way you have Say
heat lost by the uh condensate steam is
heat gained by the fluid m. CP delta t
etc. You can for a given length you'll
be able to get the exit temperature. So
okay with this we come to the end of
this section of uh internal flow where
we looked at constant wall heat flux and
constant wall temperature case. Next
module onwards we will derive very very
important expression for nasalt number
which is the non-dimensional heat
transfer in case of constant wall heat
flux and constant wall temperature case.
For this we will go back to our fluid
mechanics. Look at lamina flow through a
pipe.
Use our uh velocity distribution
equation. Derive the bulk mean
temperature. use the definition of nasal
number and get the value of Nusel number
for constant wall heat flux case. Thank
you.
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