Video summary
The lecture begins by reviewing transient heat conduction in a finite plane wall, where only one term of an infinite series is sufficient to describe temperature distribution when the Fourier number exceeds 0.2. The instructor illustrates how the temperature profile evolves over time: initially symmetric with the center line remaining at the initial temperature $T_i$ while the walls cool down due to heat removal by a fluid. Eventually, as time approaches infinity, the entire wall reaches thermal equilibrium with the surrounding environment ($T_\infty$). A significant conceptual point made is that for very high convection coefficients, such as those found in boiling or condensation scenarios where $h$ can reach 50,000 to 100,000 W/m²K, the Biot number approaches infinity. In this limit, convective resistance becomes negligible, causing the surface temperature to equalize instantly with the fluid temperature ($T_\infty = T_{wall}$). This specific case effectively transforms a convection boundary condition problem into one of constant wall temperature, simplifying the mathematical approach required for analysis.
The discussion then transitions to the concept of semi-infinite solids, defined not by infinite physical dimensions but by conditions where heat penetration is shallow relative to the total size of the object over the time period of interest. The instructor uses a practical example of a water pipe buried deep in soil in Kashmir during winter; while snow lowers the surface temperature to -10°C, the ground remains warm enough at greater depths that the pipe does not freeze for several months. Mathematically, this scenario is modeled by assuming the initial temperature $T_i$ persists indefinitely as depth ($x$) approaches infinity. To solve the resulting partial differential equation governing heat diffusion in such a medium, the lecturer introduces a similarity variable $\eta = x / \sqrt{4\alpha t}$. This transformation cleverly reduces the complex partial differential equation into an ordinary differential equation by demonstrating that temperature becomes dependent solely on this single combined variable rather than space and time independently.
Solving the transformed equation through separation of variables leads to a solution involving integrals of exponential functions, which are identified as the Error Function ($\text{erf}$). The lecture defines the error function mathematically as an integral from 0 to $Z$ of $e^{-t^2} dt$, noting its properties such as being zero at $Z=0$, approaching one as $Z \to \infty$, and representing a series expansion useful for calculations. By applying boundary conditions where the surface is held at temperature $T_s$ (at $\eta = 0$) and the interior remains at $T_i$ (as $\eta \to \infty$), the final temperature distribution formula is derived as $(T - T_s)/(T_i - T_s) = \text{erf}(\eta)$. The session concludes by emphasizing that while this solution provides a complete analytical description for constant surface temperature, future classes will extend these methods to handle other boundary conditions like constant heat flux and convection, alongside explaining the physical derivation of the similarity variable used.
Read the full video transcript
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>> So, we were into
if you just check, we were into getting
the temperature distribution of a plane
wall.
So, we did get mathematically the
temperature distribution of a plane wall
of this form and only one term series is
enough for us, we said, if the Fourier
number is greater than 0.2.
And these were the temperature
distributions. If you really look the
plane wall, which I had consciously not
shown you thus far,
is this.
So, you can see that here
this is TI and this is T infinity.
So, as
this is T equal to time T equal to zero.
As
T goes on increasing, eventually at T
tending to T infinity, T time tending to
infinity, the temperature reaches T
infinity. The plane wall temperature
reaches T infinity.
But in the meantime, the temperature
distribution, you can see that it is
symmetric, number one.
And the center line temperature is
higher compared to the
temperature at the wall because at the
wall, the heat is removed, so naturally
the temperature will be lower here.
But eventually, the temperature will
even at the center also reduces to T
infinity.
So,
but there is some time till T equal to
T2, for example, here this is T equal to
zero, this is T equal to one, T1 and
this is T equal to T2. The center line
temperature continues to be at TI.
So, till T equal to T2, the center line
temperature has not changed or has not
felt the heat taken away by this fluid.
The heat is not taken away. So, this is
a important point to note. And this is
how the temperature distribution looks
like when you plot taking that
exponential and other functions together
in a plane wall.
Okay. So, now only one issue before we
go to the semi-infinite thing which I
want to cover is that uh
the Biot number
infinite. So, I just touchingly told
that the Biot number infinity means what
does it mean? So, if you just take up
Biot number is infinite means So, let's
say Biot number is infinite. What is
Biot number? HL by K.
So, it tends to infinity
implies what does it imply? It implies
that the H is very high. Then only this
is possible for a given plane wall and a
material of given thickness. So, when
can one achieve very high H? This is
possible in case of boiling or
condensation
where the heat transfer coefficients are
can be of the order of 50,000 to 1 lakh
W per m squared K.
In that case, what happens is that the
one by Biot number
in that case, you just see one by Biot
number
that is K by HL
is zero.
That means if you have
H very high
H is infinite, then the T infinity is
equal to the T wall because there is no
convective resistance. 1 by H A is the
convective resistance. Convective
resistance is 1 by H A if you recall. In
1D conduction we said 1 by H A is
convective resistance. If H is very
large, that means this convective
resistance is
zero, that implies that T infinity equal
to T wall. That means for a Biot number
tending to infinity case, I I actually
solved for constant wall temperature
case also.
That means if you were having a plane
wall
and which is sitting at a temperature T
equal to TS.
T equal to TS and now
you want to get the temperature
distribution inside, then you can get
taking the Biot number as infinity. So,
you have not only solved the temperature
distribution for convective boundary
condition, convective heat transfer
boundary condition, which we did in the
previous classes, you have also solved
for constant
wall temperature. So, only thing the
procedure is all same for constant wall
temperature case. Only thing is that you
have to solve
you have to take the lambda one and A1
for Biot number corresponding to
infinity case and solve the rest of the
problem as same. So, that's about the
constant wall temperature case, which we
could fortuitously get from the
convective boundary condition itself.
So, with this, we can move on to what is
called as semi-infinite
medium or transient heat conduction.
Transient
heat conduction
in
semi-infinite
solids.
So, what do I mean by semi-infinite
solids?
Instead of telling it mathematically,
let me explain you this through example.
Let me explain this through an example.
Let's say
I am sitting in Kashmir
and I have in my pipe wall. This is my
pipe
in which the water is flowing.
Okay?
And this pipe is put
in soil.
This is
a
dug The soil is dug and the pipe is laid
out. So, the pipe is put in the
submerged soil. So, now the initial
temperature
is 15°
C,
let's say.
Now, this is what it is, but all of a
sudden, it starts snowing.
It starts snowing, because of which the
surface temperature
the
uh
surface temperature becomes -10° C. C.
But, now what happens? Now,
this is okay. So, then what is
semi-infinite about this?
Now, if I take How do I model this? d
squared t by
d x squared equal to 1 by alpha d t by d
t.
This is what we get.
Now, I know the initial condition. t x
comma 0 for a plane wall, if I take this
as a plane wall. This is
is t i. This is initial condition.
Okay? Now,
t at
0 comma t
that is at the surface. That is This is
my x starting. That is x equal to 0.
It is
t s. This is one of the boundary
condition. This is the snow temperature.
Let us say.
But now,
if
x is very, very lower than
very, very I have dug very deep. This x
is very large. Several tens of meters I
have dug.
Then, my temperature
at any time t
or for certain time t, it is going to be
at the t i.
Not the It won't be any different. Under
these circumstances, I can consider this
as semi-infinite.
It is a semi This is called
semi-infinite
situation.
I will elaborate this still better
through various examples after we have
solved this. But for now, this is the
physical situation where I have put a
pipe in Kashmir below the soil, and the
snow has taken over so that the surface
temperature is minus 10. But I have laid
down my pipe so deep down that my pipe
will not feel the
temperature variation caused because of
my snow or the heat transfer it doesn't
feel or how deep
I should dig how deep we need to dig so
that my
water inside the pipe does not get
frozen.
That's the practical solution. My
contractor is asking me this question.
Sir, how how
deep I should dig so that for 6 months
or for 3 months in the winter season my
water continues to be
remaining as water while it is flowing,
otherwise it will freeze down. So to
solve that, this is the problem I have
to solve. This is going to be a little
different from what we solved earlier.
That is the plain wall problem. How it
is going to be mathematically different
and what is the simple solution what I
can get instead of the complicated
solution what we got for the plain wall.
We can check that using the
semi-infinite medium. So this is what is
the semi-infinite medium. So now let us
get how do I solve this problem of
semi-infinite
medium. That is
do squared t by
do x squared equal to 1 by alpha do t by
do t.
You know that this is partial
differential
equation.
Now
I can transform this partial
differential equation into ordinary
differential equation
by using a transforming variable.
I will explain that transforming how I
did I take the transforming variable
little later through physics, but for
now you please take my words and take
this transforming variable as eta equal
to square root of X by square root of 4
alpha t. X is the space, alpha is
thermal diffusivity, t is the time. So
now, let us do this transforming of
partial differential equation to
ordinary differential equation. So how
do I do that? So do t by do t
is equal to
if I I can write this as dt by d eta
into do eta by do t.
Okay? So dt by d eta retain as it is.
Do eta by do t, if you differentiate
this with respect to t,
x by square root of 4 alpha are
constant. t to the power of minus half
will be minus half
n x to the power of n minus 1 minus half
minus 1.
So that is equal to
do t by do t
is equal to dt by d eta
into
that is negative sign into negative sign
x divided by
2 t into square root of 4 alpha t. It is
t to the power of 3 by 2 minus 3 by 2.
One I have put inside the square root
and one is outside. So this is do t by
do t.
Now let us take up do t by do x because
I need do square t by do x squared. So
first let me get do t by do x. Do t by
do x is nothing but dt by d eta
into d eta by
d do eta by do x
That is dt by d eta into do eta by do x
that is
It is differentiating with respect to x
so that is it is nothing but 1 by 4
alpha differentiation of x with the x
will be
1. So you got this as this. Now
similarly do squared t by do x squared
will be equal to d squared if we again
differentiate this you get d squared t
by d eta squared into 1 by square root
of 4 alpha t into 1 by square root of 4
alpha t you have to differentiate again
twice. So you get 1 by 4 alpha t into d
squared t by d eta squared.
Okay, so now let us substitute this
do t do squared t by do x squared these
two terms
into this equation.
So if I substitute that so I get 1 by 4
alpha t
into do squared t by do eta squared
is equal to 1 by alpha into
minus x divided by 2 t into square root
of 4 alpha t
into dt by d eta.
So this is
d squared t divided by d eta squared.
Now why it is you will realize why it
becomes ordinary differential equation
in a second you will realize. Now alpha
alpha gets cancelled t t gets cancelled.
Now
what will I get?
d squared t divided by d eta squared
equal to
-2x
divided by square root of 4 alpha t
into dt by d eta.
Now, if you recall what is eta?
Eta is x by 4 alpha t. So, this is
nothing but eta.
So, I get d squared t divided by
d eta squared is equal to -2 eta dt by d
eta.
Okay. So, now you can see that my
temperature is only a function of eta.
But, here my temperature was a function
of both x and t.
So, that is why PD That's why it was PDE
here. Now, it is ODE because my
temperature is only a function of one
variable, that is eta. But, that magic
was done through eta. How I chosen this
x by square root of 4 alpha t, I have
not explained. You will have to wait for
that. I will explain little later. Okay?
But, now let us convert the initial
conditions also into eta terms. That is
t of x {comma} 0 is equal to ti. This is
the boundary initial condition. At t
equal to 0, what happens to my eta? Eta
equal to x divided by square root of 4
alpha t. That means at t equal to 0 eta
tends to infinity.
So, at eta tending to infinity
t at eta tending to infinity
t becomes ti. This is one initial
condition.
Now, let us take up the boundary
condition. That is T at 0 {comma} T is
equal to TS.
That is at X equal to 0,
eta equal to X Y square root of 4 alpha
T, eta is also equal to 0 because X is
0. So, that means T at eta equal to 0 is
TS.
There is one more boundary condition.
T
X tending to infinity at any time T is
equal to TI.
That is the specialty of semi-infinite
medium.
That is as X tends to infinity
eta tends to infinity. That is T of eta
tending to infinity is equal to TI,
which is what we got even for
the initial condition here.
So, that means you need only two
boundary conditions because it is two
second order second order equation
second order equation. So, you have two
boundary condition that is D equation is
D squared T by D eta squared equal to
minus two eta DT by D eta.
And
T
at eta tending to infinity is TI and T
at eta equal to 0 is TS.
So, this is what we need to solve now.
So, let's see how do we solve this
equation.
Now, it is simple separation of
variables actually. We can do this. So,
what we will do is D squared T by So,
let me get started and then fresh page.
D squared T divided by d eta squared is
equal to minus 2 eta dt by d eta.
Okay? Now,
I can write this as this implies as d by
d eta of dt by d eta is equal to minus 2
eta dt by d eta.
Let
d t by d eta be p.
So, if that is the case, I get dp by d
eta
dp by d eta
is equal to minus
2 eta p.
So, I can
do this by separation of variables. That
is
dp by p
is equal to minus 2 eta d eta.
minus
2 eta d eta. Now, I can integrate it. I
have separated my variables. Now,
integrating.
If you integrate, you get log p equal to
minus 2 eta squared by 2 plus
c.
Or
dt by d
d eta. P is nothing but dt by d eta.
This is
dt by d eta. So, dt by d eta equal to c1
e to the power of minus eta squared. I
have assumed c has become c1 now. And 2
2 gets cancelled out. So, dt by d eta is
c1 e to the power of minus p squared.
Now, again, if I integrate this,
I get t equal to integrating again
with respect to eta,
T equal to I get C1
integral of e to the power of minus eta
squared d eta between the limits 0 to
eta plus C2.
So, there are two constants and I have
two boundary conditions which I have
been given. That is
T at eta equal to 0 is Ts
and T at eta tending to infinity is Ti.
Now,
T at eta equal to 0 is Ts. So, what what
will happen if I substitute this here?
That is you get
Ts equal to C1 e to the power of minus 0
d eta 0 to eta plus C2. This is 0.
So, C2 will become equal to what is
called
Ts.
Now, if I substitute this, so you get T
equal to C1 integral of 0 to eta e to
the power of minus eta squared d eta
plus Ts.
Okay? Now, we need to find C1.
So, that is
Ti equal to C1 integral of 0 to
eta e to the power of minus eta squared
d eta. Sorry, this is T.
Okay? Plus
Ts.
Now, we have another boundary condition,
T at eta tending to infinity. That is
equal to Ti.
So, Ti equal to C1
0 to infinity e to the power of minus
eta squared d eta plus TS.
Incidentally, this is what is called as
error function. We have defined what is
called as error function. I will show
you little while from now what is error
function. Error function is if you have
error function of Z equal to it is given
by 2 by root pi integral of 0 to
Z e to the power of minus T squared dt.
So, what is this error function? This
error function if you have error
function of Z, can I calculate error
function in terms of series? Yes. Error
function of Z is 2 by square root of pi
into Z minus Z cube by 3 plus Z to the
power of 5 by 10 minus Z to the power of
7 by 42 plus Z to the power of 9 divided
by 216 so on and so forth. If I have to
show you
If I have to show you the graph, how
does it uh the figure how does it look
like?
So, this is how the error function looks
like. So, the error function definition
is encountered in
in integrating normal distribution
curves by the Gaussian distribution. So,
error function Z is 2 by square root of
pi 0 to Z e to the power of minus T
squared dt. If you plot that, this is
how it looks. This is error function and
this is Z. So, this is the series
function. So, what is interesting here
is that error function of 0, you can see
that it is 0 and error function of
infinity
is 1. Error function of infinity is one.
And it's an odd function. That means
error function of minus Z is equal to
minus of error function of Z.
So, error function
So, let us write few
conclusions of error function. That is
error function of
zero is zero. You can see that here. Is
zero. And error function of infinity is
one.
And error function of
minus Z is equal to error function of
minus of error function of Z.
Okay? That is And also error function of
Z plus
error function
complementary error function of Z is
equal to one. So, what is this? This is
what is called as complementary
error function. We are going to use this
little later. So, that's why I am
putting all of this. So, to imagine
error function is just a series equation
like this. Now, let us get back and get
our C1.
So, if you get that
C1, so you get TI equal to this is the
error function. So, if it is infinity,
so what do I get? This is C1.
into. So, the error function is this.
So, this becomes the error function. So,
I can write e to the power of this
becomes error function of
That is you get this as zero to infinity
C1. Zero to infinity e to the power of
minus
Sorry, let me write in one shot. C1
equal to
TI minus TI
TS divided by 0 to infinity e to the
power of minus eta squared d eta.
Now,
this being equal to 1 for infinity, so
this error function of infinity is equal
to 1. That means that is equal to 2 by
square root of pi integral of 0 to z e
to the power of minus t squared dt. So,
this becomes equal to
square root of pi by 2. So, that is TI
by TS
divided by square root of pi by 2. So,
that is C1 equal to 2 into TI minus TS
divided by square root of pi.
Okay? So, you we got C1, so we got C2.
So, now we have gotten the temperature
distribution. If I substitute that here,
I can get the temperature distribution.
So, my temperature distribution is
T equal to C1 integral of 0 to eta e to
the power of minus eta squared d eta
plus C2. So, C2 is TS and C1 is 2 into
TI minus TS divided by square root of
pi.
So, T equal to 2 into TI minus TS
divided by square root of pi
into 0 to eta e to the power of minus
eta squared d eta plus TS. So, T minus
of temperature at any time T minus T S
divided by T I minus T S
Sorry, temperature at any eta this is.
is equal to
2 in 2 by square root of pi.
This is nothing but error function.
The whole of this this is 0 to eta.
Sorry, e to the power of minus eta
square d eta. This is nothing but error
function of eta.
So,
T minus T of eta minus T S divided by T
I minus T S is nothing but error
function of eta. So,
if I know that T, I know T I, I know T S
at any eta, I need to get the T by
getting the error function. How do I get
the error function? Actually, it is
calculated and tabulated. The error
function is used calculated using this
function, but you don't have to
calculate every time. It is calculated
and kept. So, if this is your eta, the
error function of eta is calculated and
kept. So, that's about the temperature
distribution. In the next class, I am
going to teach you about the surface
heat flux and also how at other boundary
conditions. So, we are solving it for
constant temperature surface
temperature. How can it How can What
will be the solution for constant heat
flux and surface convection will be
covered in the next class. And also, we
will solve this problem. And another
question which is pending is how on
earth I came to know that eta equal to X
by square root of 4 alpha T. So, these
two questions I will be addressing them
in the next class. Thank you.
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