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MORGAN BINGGELI: Hi, everyone.

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My name is Morgan Binggeli.

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I'm a first year masters student
at EPFL in material science

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and engineering, and
I'm going to present

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to you this video about
heat transfer in a material.

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This video start with
a little introduction

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where I will give you
some definitions which

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will be useful
for heat transfer,

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and where I will talk a
bit about heat equations.

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Then I'm going to present to you
some concrete examples that you

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can meet in your everyday life.

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In order to be able to see the
next example in a proper way,

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we need to give
some definitions.

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In this video we're going to say
that heat is a form of energy.

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The temperature is a
measurable manifestation

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of the stored heat Instabilities
in which different temperatures

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are in contact--

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a heat transfer occurs,
transferring the heat

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from the warmer to
the colder body.

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Several heat transfer
modes exists.

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Through heat conduction,
which corresponds

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to the heat exchanged between
two points of a motionless

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and opaque solid,
liquid, or gas.

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Through convention, which
corresponds to a heat exchange

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between the wall and a
fluid, with a heat transfer

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done by the moving fluid.

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Finally, the
radiation corresponds

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to a heat exchange between
two walls, separated

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by a transparent environment.

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As we're going to talk about
heat transfer in this video,

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it's important to understand
our heat equations

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in this kind of phenomena.

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First, the heat
conduction law, which

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defines the diffusive
thermal flux and function

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of the temperature gradient.

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And this relation
is linearly governed

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by the thermal conduction
coefficient created here.

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Then, the heat equation, which
has a time-dependent term,

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a speed-dependent term--

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which correspond to the
advection phenomena.

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A divergence term, here, of
the diffusive thermal flux.

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And finally, a heat system
which corresponds often

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to chemical reactions which are
either esoteric or endothermic.

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Combining both equations,
we get this final relation,

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which defines heat transfer
and heat phenomena.

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We're now going to
apply these relations

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to some concrete examples.

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As a first example, a simple
wall of a house with thickness,

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e, is chosen.

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Its thermal state is stable,
and there is no [INAUDIBLE]..

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A thin wall is considered,
with an unidirectional heat

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flux along x-axis.

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The temperatures on both sides
of the wall are different.

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T1, here, is lower
than T2, here.

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The temperature profile
in the wall is researched.

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To solve this problem,
it is adept to use

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the Cartesian system.

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The temperature only
changes along x-axis,

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and the nonhomogenous
distribution of the temperature

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is weighted in the
other directions.

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The conditions of
this problem allow

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to simplify the heat
equations presented before,

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which becomes such an equation.

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As boundary conditions,
we define the temperature

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in x equals 0 to T1, here.

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And in x equal e to T2, here.

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We then defined the system
and try to solve it.

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You can see here
that the solution is

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dependent on e, the thickness.

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To be able to have a general
solution, nondependent

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of this thickness, a
variable change is done.

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The temperature is not in
function of x, anymore--

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as it is here--

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but in function of
the ratio x over e,

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which varies between 0 and 1.

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The differential system of
equations and the solution

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are as follows.

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Here, we define the system.

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And here, we try to solve it.

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The solutions look like this.

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Changing the disposition
of the equation,

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it's possible to get
this kind of solution.

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It's then possible to generalize
this equation to every case,

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doing the following
transformation, here.

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So we can now see
that our solution

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is only dependent on
an extra e variable.

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Thus, it's possible to
plot the temperature

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profile for this case.

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It looks like this.

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So we have a linear
behavior between T1, here,

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and T2, here, along the x-axis.

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It may be easier to see the
temperature distribution

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in the wall.

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Here we can clearly see
that, if it's our wall,

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we have a linear gradient
of the temperature,

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with the minimum at T1, and,
here, our maximum at T2.

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That's all for
this first example.

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As a second example, a concrete
wall with a thickness, e,

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is chosen.

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This concrete wall is hardening
due to an exothermic chemical

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reaction, like a direction,
for the concrete.

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The temperature is the same
at both sides of the wall,

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and is equal to Tw.

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This case is shown as a
stable and motionless case.

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To solve this problem,
it is adept to use

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the Cartesian system.

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The temperature only
changes along the x-axis,

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and an homogeneous
distribution of the temperature

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is weighted in the
other directions.

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The conditions of
this problem allow

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to simplify the heat equation
presented before, which

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becomes this kind of equation.

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At the boundary conditions,
we have the temperature

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defined in x equals 0
and x equal Te, as Tw.

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Again, we define the
differential equation system

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and it's boundary conditions,
and try to solve it.

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To get the general
solution, usable

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for a large amount
of cases, it's

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much likely to be easier
to [INAUDIBLE] results.

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Thus, a variable change is done.

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The temperature is not
in function of x anymore,

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as before, but in function
of the ratio x over e--

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which varies between 0 and 1,
as it was explained before.

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The differential system of
equations and the solution

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are as following.

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So the differential system
of equations is this one,

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and its solution
looks like this.

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Furthermore, all temps that
are independent of x over e

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are placed to the left
side of the equals sign.

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Thus, the temperature
profile looks like this.

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We can clearly see a symmetrical
behavior in this solution,

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with a maximum in the
center of the piece, where

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the temperature is higher due
to the exothermic phenomena.

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The minimum are at
the faces, where

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the temperature is equal to Tw.

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We can also see the
temperature distribution

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in the wall as following.

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We can clearly see this maximum
in the center of the piece,

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and the minima at
the side of the wall.

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As a last example,
a thin steel bar

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is considered during a
continuous thermal treatment

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process.

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The bar is extracted
from an oven,

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heated at a temperature
of T0, and then quenched

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after L, distance, in
a water bath, which

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is at a TL temperature.

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The temperature
profile of the bar,

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between the exit of the oven and
the entrance of the water bath

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is researched when a steady
state is established.

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To solve this problem,
it is adept to use

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the Cartesian system.

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As the bar is still,
a uniform temperature

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is assumed in its
translucent section.

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Which means, the
temperature will only

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depend on the exposition.

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The conditions of
this problem allow

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to simplify the heat
equation presented before,

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which becomes this equation.

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As boundary conditions, we
have that the temperature in x

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equals 0, here, is equal to T0.

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And the temperature in x
equal L is equal to TL.

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The differential equation
system is the following,

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and its solution is as follows.

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This looks quite
complicated, but we

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can simplify a bit knowing
the factor rho times Cp

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over K, is one of alpha--

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where alpha is the thermal
diffusivity coefficient

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of the material.

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We still have this solution.

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This solution also
looks complicated.

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However, it appears
simpler rewriting it

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in the following way.

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As for the previous example, a
general solution is researched.

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In this case, this
solution looks like this.

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We can only add that.

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We still have this
solution, here.

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And knowing the ratio vL
over alpha corresponds

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to the adimensional
Péclet number--

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which defines the ratio
between advection and diffusion

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phenomena for our process--

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it's possible to simplify a bit.

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Here, the Péclet number is put
at the denometer, and here,

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at the numerator.

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Finally, in the same way
as the previous cases,

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a variable change is done
from replacing x over L

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with the variable x to L.

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It's possible to try to
understand how the Péclet

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number works, varying the
temperature profile in function

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of it.

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Here, we have the temperature
profile with a low Péclet

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number.

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We see that it has a mostly
linear behavior as we

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have for the first example.

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However, when we increase
the Péclet number,

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we see that it has increased
the curvature of our curve.

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And when we arrive at
a big Péclet number,

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we almost have a
vertical line-- here.

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That's due to the
speed of the bar.

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It's possible to show this
result with the graphical

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solution of the temperature
profile for a different Péclet

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number.

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We define a list
of Péclet numbers,

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apply the solution on it.

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The graphical solution
looks like this.

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We can see that the solution
for the low Péclet numbers are

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quite linear, and that it has
a bigger more and more curve

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as the Péclet number increases--

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the number you end with, here.

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Finally, it's also possible to
present a solution in a wall

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allowing a variable
Péclet number.

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Here, we can clearly see that
the gradient is linear when

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the Péclet number is low.

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But if we increase it, we
see that the differential

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of the temperature is almost
on the side of the bar.

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So that's all for this
video about heat transfer

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in the material.

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Here are my references.

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Thank you for watching.