In order to get to the bottom of things, we will now look at the influence of the thickness of the component separating the two reservoirs. For this, as a first step we will assume that this is a homogeneous single-layer component, i.e. the entire thickness is filled with a single material which also has the same density everywhere. It quickly turns out that for most materials the measured U-value of the layer is inversely proportional to the layer thickness, at least below the boundary thickness $d_{Grenz}$. Again, this is not surprising if we think of the vibrational energy of the molecules being passed on as a model for thermal conductivity. The longer the distance is across which the vibrational energy must be transferred, the greater the thermal resistance will be (and that is $R:=1/U$). There is a proportionality factor which mainly depends only on the type of material - and this factor is called the thermal conductivity, the abbreviation for which is the Greek letter λ. Therefore:

${\displaystyle \hspace{2cm} U= \frac{\lambda}{d} \hspace{6cm} [\lambda 1]}$

If the thickness is equal to a unit of length (this is 1 m in the International System of Units SI), then the thermal conductivity simply indicates the U-value which the material would have with a thickness of 1 m. Of course, in practice this thickness is huge, that is why many thermal conductivities seem so “small”. The thermal conductivity is stated with the unit of measurement W/(mK). The thermal conductivities of thousands of materials have been measured and documented because this value is needed all the time for technical systems¹⁾. If we want to achieve low heat losses then it is helpful to strive for low U-values in particular (there's nothing we can do about the temperature difference and the areas cannot be reduced below a certain size for the given requirements). Low U-values can be achieved with large thicknesses on the one hand²⁾, and through low thermal conductivities on the other hand; in fact, the latter are very different for various materials: while these can be several 100 watts per metre and kelvin for metals (copper: λ=380 W/(mK) ), we come down to 0.0095 W/(mK) with a vertical layer of krypton gas ³⁾.

This table clearly shows that:

• Metals have a very high conductivity. Practical use: if heat has to be dissipated quickly, then e.g. a cooling element/heat sink of aluminium is used.
• However, conventional (heavy) mineral building materials also conduct heat well. A “traditional” masonry wall therefore has high heat losses.
• Thermal conductivities usually decrease with declining gross densities. There's a simple reason for this: a lot of air is trapped in the lighter materials.
• Gases have the lowest thermal conductivities, i.e. heat transfer via thermal conduction. But this would only work for as long as the gases are still and do not transport⁸⁾ heat in a flow; this happens very fast due to a change in the density and thermal buoyancy. For the gas to exhibit an insulation effect despite this, the flow resistance must be large enough. This is the case with small layer thicknesses - or by building up flow impedances/creating impedances in the flow, such as the cells of a foam insulation material or e.g. jute fibres⁹⁾.
• Conventional thermal insulation materials are basically nothing other than air packed in tiny portions.
• The larger the mass of the individual molecules in a gas is, the smaller its thermal conductivity will be. This is paradoxical only at the first glance, because the heavier gases vibrate more slowly at the same temperature on account of the inertia of their molecules. In practice, this is why the intermediate spaces between glass panes are not filled with air in industry and heavier gases like argon or even krypton are used instead ¹⁰⁾.
• Yet another “curiosity” of thermal conduction in gases is the fact that their thermal conductivities are practically independent from pressure, down to extremely low gross densities and gas pressures. This fact is easily explained by the kinetic model of gas: with lower densities the molecules have a longer free path length, so until the next collision, they transport the heat further. This compensates for the smaller number of molecules. Only if the distance $d$ becomes so small that the molecules are just plopping back and forth between the two walls, the thermal conductivity starts to decrease with further reduction of the density. We make use of this for creating what is called “vacuum insulation” ¹¹⁾.
• Note: thermal radiation plays a more significant role in heat transfer in such an intermediate space, especially in the case of gas fillings ¹²⁾. We will deal with this elsewhere ¹³⁾.
• Building envelope areas with reasonable component thicknesses is only possible if the essential insulating effect comes from a good thermal insulation material.

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