MORGAN BINGGELI: Hi, everyone.
My name is Morgan Binggeli.
I'm a first year masters student at EPFL in material science
and engineering, and I'm going to present
to you this video about heat transfer in a material.
This video start with a little introduction
where I will give you some definitions which
will be useful for heat transfer,
and where I will talk a bit about heat equations.
Then I'm going to present to you some concrete examples that you
can meet in your everyday life.
In order to be able to see the next example in a proper way,
we need to give some definitions.
In this video we're going to say that heat is a form of energy.
The temperature is a measurable manifestation
of the stored heat Instabilities in which different temperatures
are in contact--
a heat transfer occurs, transferring the heat
from the warmer to the colder body.
Several heat transfer modes exists.
Through heat conduction, which corresponds
to the heat exchanged between two points of a motionless
and opaque solid, liquid, or gas.
Through convention, which corresponds to a heat exchange
between the wall and a fluid, with a heat transfer
done by the moving fluid.
Finally, the radiation corresponds
to a heat exchange between two walls, separated
by a transparent environment.
As we're going to talk about heat transfer in this video,
it's important to understand our heat equations
in this kind of phenomena.
First, the heat conduction law, which
defines the diffusive thermal flux and function
of the temperature gradient.
And this relation is linearly governed
by the thermal conduction coefficient created here.
Then, the heat equation, which has a time-dependent term,
a speed-dependent term--
which correspond to the advection phenomena.
A divergence term, here, of the diffusive thermal flux.
And finally, a heat system which corresponds often
to chemical reactions which are either esoteric or endothermic.
Combining both equations, we get this final relation,
which defines heat transfer and heat phenomena.
We're now going to apply these relations
to some concrete examples.
As a first example, a simple wall of a house with thickness,
e, is chosen.
Its thermal state is stable, and there is no [INAUDIBLE]..
A thin wall is considered, with an unidirectional heat
flux along x-axis.
The temperatures on both sides of the wall are different.
T1, here, is lower than T2, here.
The temperature profile in the wall is researched.
To solve this problem, it is adept to use
the Cartesian system.
The temperature only changes along x-axis,
and the nonhomogenous distribution of the temperature
is weighted in the other directions.
The conditions of this problem allow
to simplify the heat equations presented before,
which becomes such an equation.
As boundary conditions, we define the temperature
in x equals 0 to T1, here.
And in x equal e to T2, here.
We then defined the system and try to solve it.
You can see here that the solution is
dependent on e, the thickness.
To be able to have a general solution, nondependent
of this thickness, a variable change is done.
The temperature is not in function of x, anymore--
as it is here--
but in function of the ratio x over e,
which varies between 0 and 1.
The differential system of equations and the solution
are as follows.
Here, we define the system.
And here, we try to solve it.
The solutions look like this.
Changing the disposition of the equation,
it's possible to get this kind of solution.
It's then possible to generalize this equation to every case,
doing the following transformation, here.
So we can now see that our solution
is only dependent on an extra e variable.
Thus, it's possible to plot the temperature
profile for this case.
It looks like this.
So we have a linear behavior between T1, here,
and T2, here, along the x-axis.
It may be easier to see the temperature distribution
in the wall.
Here we can clearly see that, if it's our wall,
we have a linear gradient of the temperature,
with the minimum at T1, and, here, our maximum at T2.
That's all for this first example.
As a second example, a concrete wall with a thickness, e,
is chosen.
This concrete wall is hardening due to an exothermic chemical
reaction, like a direction, for the concrete.
The temperature is the same at both sides of the wall,
and is equal to Tw.
This case is shown as a stable and motionless case.
To solve this problem, it is adept to use
the Cartesian system.
The temperature only changes along the x-axis,
and an homogeneous distribution of the temperature
is weighted in the other directions.
The conditions of this problem allow
to simplify the heat equation presented before, which
becomes this kind of equation.
At the boundary conditions, we have the temperature
defined in x equals 0 and x equal Te, as Tw.
Again, we define the differential equation system
and it's boundary conditions, and try to solve it.
To get the general solution, usable
for a large amount of cases, it's
much likely to be easier to [INAUDIBLE] results.
Thus, a variable change is done.
The temperature is not in function of x anymore,
as before, but in function of the ratio x over e--
which varies between 0 and 1, as it was explained before.
The differential system of equations and the solution
are as following.
So the differential system of equations is this one,
and its solution looks like this.
Furthermore, all temps that are independent of x over e
are placed to the left side of the equals sign.
Thus, the temperature profile looks like this.
We can clearly see a symmetrical behavior in this solution,
with a maximum in the center of the piece, where
the temperature is higher due to the exothermic phenomena.
The minimum are at the faces, where
the temperature is equal to Tw.
We can also see the temperature distribution
in the wall as following.
We can clearly see this maximum in the center of the piece,
and the minima at the side of the wall.
As a last example, a thin steel bar
is considered during a continuous thermal treatment
process.
The bar is extracted from an oven,
heated at a temperature of T0, and then quenched
after L, distance, in a water bath, which
is at a TL temperature.
The temperature profile of the bar,
between the exit of the oven and the entrance of the water bath
is researched when a steady state is established.
To solve this problem, it is adept to use
the Cartesian system.
As the bar is still, a uniform temperature
is assumed in its translucent section.
Which means, the temperature will only
depend on the exposition.
The conditions of this problem allow
to simplify the heat equation presented before,
which becomes this equation.
As boundary conditions, we have that the temperature in x
equals 0, here, is equal to T0.
And the temperature in x equal L is equal to TL.
The differential equation system is the following,
and its solution is as follows.
This looks quite complicated, but we
can simplify a bit knowing the factor rho times Cp
over K, is one of alpha--
where alpha is the thermal diffusivity coefficient
of the material.
We still have this solution.
This solution also looks complicated.
However, it appears simpler rewriting it
in the following way.
As for the previous example, a general solution is researched.
In this case, this solution looks like this.
We can only add that.
We still have this solution, here.
And knowing the ratio vL over alpha corresponds
to the adimensional Péclet number--
which defines the ratio between advection and diffusion
phenomena for our process--
it's possible to simplify a bit.
Here, the Péclet number is put at the denometer, and here,
at the numerator.
Finally, in the same way as the previous cases,
a variable change is done from replacing x over L
with the variable x to L.
It's possible to try to understand how the Péclet
number works, varying the temperature profile in function
of it.
Here, we have the temperature profile with a low Péclet
number.
We see that it has a mostly linear behavior as we
have for the first example.
However, when we increase the Péclet number,
we see that it has increased the curvature of our curve.
And when we arrive at a big Péclet number,
we almost have a vertical line-- here.
That's due to the speed of the bar.
It's possible to show this result with the graphical
solution of the temperature profile for a different Péclet
number.
We define a list of Péclet numbers,
apply the solution on it.
The graphical solution looks like this.
We can see that the solution for the low Péclet numbers are
quite linear, and that it has a bigger more and more curve
as the Péclet number increases--
the number you end with, here.
Finally, it's also possible to present a solution in a wall
allowing a variable Péclet number.
Here, we can clearly see that the gradient is linear when
the Péclet number is low.
But if we increase it, we see that the differential
of the temperature is almost on the side of the bar.
So that's all for this video about heat transfer
in the material.
Here are my references.
Thank you for watching.