11

The thermal conductivity coefficients of some gasses in units of W/(m•K) are helium 0.142, argon 0.016, air 0.026, methane 0.03, propane 0.015, bromine 0.004, and steam 0.018. These are all rather similar except for helium and bromine. Using the kinetic theory of gasses the thermal conductivity coefficient is given by $ \kappa \sim n\bar c\lambda C_V$ where ...


3

If the heat transfer in fluid 2 is fast, the thickness of the boundary layer wall - fluid 2 is very small (the 'not so linear' part of the temperature profile on that side) One can then approximate that the temperature of fluid 2 directly at the wall is equivalent to the bulk temperature. So the temperature profile through the wall would be steeper, ...


3

Newton's law of cooling is $$q = A h (T_0 - T_S )$$ With the area of the sphere being $$A = 4 \pi R^2$$


2

It would be nice to have a full chromatogram rather than a cropped one to see how other peaks look like. The phenomenon of odd behavior of hydrogen with helium as carrier gas with TCD detector in quite well known, but I am not sure if it has been investigated in detail in the literature or not. Hydrogen has higher thermal conductivity than helium. Ideally ...


2

This is the heat equation analogue to Ohm's law, in which resistances are additive (in series), and hence the inverse conductances are additive. The way to reason is that the full equation (1) for the heat conductivity (1) contains three terms: one for the interior of the tube, one for the tube wall, and a third for the exterior. When you look at the ...


2

Starting from the following equation (the one dimensional Fourier law) for heat flux $q$: $$q=-k\frac{\mathrm{d}T}{\mathrm{d}x}$$ where $k$ is the material's thermal conductivity, I obtained the following relations: $$|q_{2\to1}|=q_{1}=k_1\frac{\Delta{T_1}}{L}$$ $$|q_{2\to3}|=q_{3}=k_3\frac{\Delta{T_3}}{L}$$ $$|q_{2\to1}|=k_2\frac{\Delta{T_{12}}}{L_{12}}...


2

Your first line of thinking to part a) is correct. For conductive heat transfer through rectangular layers, the heat flux needs to be constant and equivalent through all layers for steady state, or you would have a 'build up' of heat at between one layer. This would locally increase the temperature, and we wouldn't be at steady state. For b) your second ...


1

This reference may be a good place to get started: Thermoelectric power and thermal conductivity in the silver-gold alloy system from 3-300°K https://doi.org/10.1080/14786437008228219


1

It's not really the same vanadium dioxide. The 67°C temperature corresponds to a phase change. Compare with the phase change in solid tin.


1

No, the temperature gradient is not uniform across all three layers. To see why, start from the Fourier law for heat flow, shown in my solution to this problem. Since at steady state the heat flux $q$ must be equal for all layers, we have that $$q_{1}=q_{2}=q_{3}$$ And since the layers are equal in thickness, $$ k_1\Delta{T_1}=k_2\Delta{T_2}=k_3\...


1

Hydrogen is the only element with thermal conductivity greater than helium ($0.182$ vs $\pu{0.151 W\,m^{-1}K^{-1}}$ at $\pu{25 ^{\circ}C}$) as M. Farooq pointed out (cf., thermal conductivity of $\ce{N2}$ is $\pu{0.026 W\,m^{-1}K^{-1}}$ at $\pu{25 ^{\circ}C}$). The mixtures of $\ce{H2}$ in $\ce{He}$ at moderate temperatures exhibit varying thermal ...


1

You are wrong to say that graphite has free electrons. The electrons are bound to sheets of aromatic rings--delocalized but not free. In electrical conduction, the electron pairs are able to jump from bond to bond on the sheet and between sheets. It is worth noting that graphite has anisotropic conduction where between sheets conduction is lower than ...


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