## Thermal Analysis of the System

In order to determine the existence of the thermal gradient in the sample, which is the main purpose of this thesis, a thermal simulation is needed. In the following chapters the thermal distribution of the system components will be studied for a characteristic case in which the sample has a mean temperature of 455ºC.

### Thermal Distribution of the System

The images attached below show the temperature distribution of the whole system sectioned longitudinally. The heating cylinder is, obviously, the hottest element in the system, followed by the sample and the upper components of the support. As the distance from the center of the system (sample and the center of the heating cylinder) increases, the temperature decreases noticeably all the way to the bottom of the base, which is the coldest point in the system.

Next the thermal distribution of the key components that make up the system, that is to say, the ones that determine the magnitude of the thermal gradient in the sample, will be discussed in more detail.

#### Heating Cylinder

The heating cylinder is the component that brings thermal energy to the system, therefore, the thermal distribution that takes place in itself will have an important effect over the temperature distribution in the sample.

If the thermal distribution of the heating cylinder is studied one can notice the existence of some thermal areas clearly differentiated. These areas are:

• Central Area: The area that covers the sample, as a consequence, part of the thermal power delivered by the cylinder in this region goes to the sample. The external surface of the heater, however, remains uncovered by the shields, so that in this region both the inner and outer faces of the cylinder are transferring radiative heat, favouring the decrease of the temperature in this area.
• Upper Area: This area is covered by the shields, therefore, the temperature rises because part of the radiation delivered by the external surface comes back to the cylinder. Its internal surface is not covering the sample at this point, it is empty, being the internal points overheated by themselves. Both factors cause a considerable rise of the temperature in this area.
• Lower Area: Like the previous area, it is covered by the shields. The difference, however, is found in the internal surface that is covering the upper part of the sample support. This fact favours a decrease of the temperature since part of the heat delivered by the cylinder is absorbed by the support.

#### Sample support

The support of the sample is a key component when it comes to find the temperature profile of the sample. This is because it is the component that joins the sample with the container and here is where the thermal losses by conduction of the sample occur. The magnitude of these losses is defined by the geometry and thermal conductivity of the materials of the components that make up the support.

It can be noticed that a big temperature difference exists along the support. It can indicate that, either the flux of heat losses is high, or the thermal conductivity of the alumina tube is low.

The magnitude of the thermal losses by conduction between the sample and the support is defined by the thermal conductivity of the materials of the support components (boron nitride, alumina and steel) and its own geometry.

#### Sample

The temperature distribution of the sample is the variable to study in this thesis, since an excessively big temperature difference in the sample would limit the validity of the results obtained in the neutron diffraction experiments.

It is seen that the sample shows a temperature difference of around 10ºC between the top and the bottom. It can be noticed, also, that the temperature difference between two adjacent points is bigger in the lower and middle areas than it is in the upper area. In order to explain the variation of the slope in the temperature profile of the sample it is necessary to resort to the equation of heat transfer by conduction between two points:

$q=-\dfrac{\lambda}{L} \cdot A \cdot \Delta T$

where q [W] is the thermal power transferred, $\lambda$ $[W \cdot m^{-1} \cdot K^{-1}]$, the thermal conductivity, A [m²], the area, L [m] is the distance between both points and $\Delta T [K]$, the temperature difference between them.

The thermal power transferred by conduction between
two points is proportional to the ΔT between them.

Since the vertical distance between the cell centers of two adjacent cells is constant for all the cells in the sample, like it is the heat transfer area between them, then $\lambda , A$ and L are constant between any couple of cells. Therefore, the power transmitted between one and another is proportional to the ΔT between them. Taking this observation into account and analyzing the plot of the temperature profile of the sample, it can be seen that the cells from the top of the sample are delivering less heat to the lower cell than the cells from the bottom and middle areas of the sample.

This phenomenon has a simple explanation: any of the external cells of the sample, not in the ends, is absorbing a radiative heat flux from the outside and a thermal flux by conduction coming from the upper cell, simultaneously it is delivering all this energy by conduction to the lower cell (it also delivers a small amount of energy to the inner cell, but this amount is insignificant). This way the temperature profile is very smooth in the top of the sample but its slope gets more pronounced in the lower areas, because more energy is transferred between adjacent cells. Finally, in the lower end of the sample a decrease in the slope can be noticed, this is because the lower end is covered by the boron nitride support so the cells in the lower end of the sample do not receive the radiative heat flux from the outside.