Last edited: 2024-10-29 14:31:52

**The Angstrom Method is a method to determine the thermal conductivity or the thermal diffusivity of different materials. The method builds upon sending temperature pulses from a reservoir and measuring the temperature at different points in the material.**

The situation is determined by the heat equation,

$\frac{\partial T}{\partial t}(\mathbf{r},t) = \alpha \Delta T(\mathbf{r},t) - \epsilon T(\mathbf{r},t),$where $T(\mathbf{r},t)$ is the temperature as a function of position $\mathbf{r}$ and time $t$, $\alpha$ is the thermal diffusivity of the material and $-\epsilon T$ represents the heat losses to the ambient air. In the case of the material being an oblong rod, the situation can be approximated as one-dimensional, which gives the following formulation of the heat equation

$\frac{\partial T}{\partial t}(x,t) = \alpha \frac{\partial^2 T}{\partial x^2}(x,t) - \epsilon T(x,t).$Furthermore, $\alpha = \frac{\kappa}{c \rho}$, where $\kappa$ is the thermal conductivity, $s$ is the speific heat capacity and $\rho$ is the density, as well as $\epsilon = \frac{R O}{S c \rho}$ where $R$ is an emission constant, $O$ is the circumference of the rod and $S$ the cross-sectional area of the rod, meaning

$\frac{\partial T}{\partial t}(x,t) = \frac{\kappa}{c \rho} \frac{\partial^2 T}{\partial x^2}(x,t) - \frac{R O}{S c \rho} T(x,t)$According to the Angstrom method, a heater shall now be connected to one of the rod's ends, say at $x=0$, at which it works as a reservoir securing the boundary condition $T(0,t) = T_0 \cos \omega t$ for any chosen temperature $T_0$ och ''angular velocity'' $\omega$. Due to diffusion and emission, the temperature $T$ will not only vary sinusoidal with $t$ and $x$ but also exponentally decrease with $x$. Therefore, a solution with the form

$T(x,t) = A e^{-ax} \cos(\omega t - b x),$where $A$, $a$ (probably $>0$) and $b$ are constants, will be assumed. If this is put into the heat equation for the oblong rod then

$-\omega A e^{-a x} \sin(\omega t - b x) = \frac{\kappa}{c \rho} A e^{-a x} \big((a^2-b^2) \cos(\omega t - b x) - 2 a b \sin(\omega t - b x)\big) - \frac{R O}{S c \rho} A e^{-a x} \cos(\omega t - b x),$is required, which if the exponential factor $A e^{-a x}$ is canceled, gives two equations in $\sin$ and in $\cos$

$\sin \implies -\omega = 2 a b \frac{\kappa}{c \rho},$ $\cos \implies \frac{\kappa}{c \rho} (a^2-b^2) - \frac{R O}{S c \rho} = 0.$Furthermore, the Angstrom method is based on measuring the temperature, $T_1$ and $T_2$, at two points, $x_1$ and $x_2$ (with $x_2>x_1$ for simplicity), and by comparing the amplitudes, say $A_1$ and $A_2$ respectively, at those points. With the nature of the temperature function, it follows that

$\frac{T_2}{T_1} = \frac{A_2}{A_1} = e^{-a(x_2-x_1)},$or

$a = \frac{\log(\frac{A_1}{A_2})}{x_2-x_1},$and by comparing the phases, say $\varphi_1$ and $\varphi_2$ respectivaly, it follows that

$\varphi_2-\varphi_1 = b(x_2-x_1),$or

$b = \frac{\varphi_2-\varphi_1}{x_2-x_1}.$and with these equations, $a$ and $b$ can be eliminated, which gives the following expression for the thermal conductivity $\kappa$

$\kappa = \frac{c \rho \omega (x_2-x_1)^2}{2 (\varphi_1-\varphi_2) \log(\frac{A_1}{A_2})}.$Using the above equation for $\kappa$ we can insert that into the equation for the thermal diffusivity

$\alpha = \frac{\kappa}{c \rho} = \frac{\omega (x_2-x_1)^2}{2 (\varphi_1-\varphi_2) \log(\frac{A_1}{A_2})}.$So that is how you determine the thermal diffusivity and thermal conductivity of a material using the Angstrom Method.

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