Determining Thermal Conductivity Using The Angstrom Method

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

Thumbnail

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 Heat Equation

The situation is determined by the heat equation,

Tt(r,t)=αΔT(r,t)ϵT(r,t), \frac{\partial T}{\partial t}(\mathbf{r},t) = \alpha \Delta T(\mathbf{r},t) - \epsilon T(\mathbf{r},t),

where T(r,t)T(\mathbf{r},t) is the temperature as a function of position r\mathbf{r} and time tt, α\alpha is the thermal diffusivity of the material and ϵT-\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

Tt(x,t)=α2Tx2(x,t)ϵT(x,t). \frac{\partial T}{\partial t}(x,t) = \alpha \frac{\partial^2 T}{\partial x^2}(x,t) - \epsilon T(x,t).

Furthermore, α=κcρ\alpha = \frac{\kappa}{c \rho}, where κ\kappa is the thermal conductivity, ss is the speific heat capacity and ρ\rho is the density, as well as ϵ=ROScρ\epsilon = \frac{R O}{S c \rho} where RR is an emission constant, OO is the circumference of the rod and SS the cross-sectional area of the rod, meaning

Tt(x,t)=κcρ2Tx2(x,t)ROScρT(x,t) \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=0x=0, at which it works as a reservoir securing the boundary condition T(0,t)=T0cosωtT(0,t) = T_0 \cos \omega t for any chosen temperature T0T_0 och ''angular velocity'' ω\omega. Due to diffusion and emission, the temperature TT will not only vary sinusoidal with tt and xx but also exponentally decrease with xx. Therefore, a solution with the form

T(x,t)=Aeaxcos(ωtbx), T(x,t) = A e^{-ax} \cos(\omega t - b x),

Using Two Measuring Points to Determine κ\kappa

where AA, aa (probably >0>0) and bb are constants, will be assumed. If this is put into the heat equation for the oblong rod then

ωAeaxsin(ωtbx)=κcρAeax((a2b2)cos(ωtbx)2absin(ωtbx))ROScρAeaxcos(ωtbx), -\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 AeaxA e^{-a x} is canceled, gives two equations in sin\sin and in cos\cos

sin    ω=2abκcρ, \sin \implies -\omega = 2 a b \frac{\kappa}{c \rho}, cos    κcρ(a2b2)ROScρ=0. \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, T1T_1 and T2T_2, at two points, x1x_1 and x2x_2 (with x2>x1x_2>x_1 for simplicity), and by comparing the amplitudes, say A1A_1 and A2A_2 respectively, at those points. With the nature of the temperature function, it follows that

T2T1=A2A1=ea(x2x1), \frac{T_2}{T_1} = \frac{A_2}{A_1} = e^{-a(x_2-x_1)},

or

a=log(A1A2)x2x1, a = \frac{\log(\frac{A_1}{A_2})}{x_2-x_1},

and by comparing the phases, say φ1\varphi_1 and φ2\varphi_2 respectivaly, it follows that

φ2φ1=b(x2x1), \varphi_2-\varphi_1 = b(x_2-x_1),

or

b=φ2φ1x2x1. b = \frac{\varphi_2-\varphi_1}{x_2-x_1}.

and with these equations, aa and bb can be eliminated, which gives the following expression for the thermal conductivity κ\kappa

κ=cρω(x2x1)22(φ1φ2)log(A1A2). \kappa = \frac{c \rho \omega (x_2-x_1)^2}{2 (\varphi_1-\varphi_2) \log(\frac{A_1}{A_2})}.

Determining Thermal Diffusivity

Using the above equation for κ\kappa we can insert that into the equation for the thermal diffusivity

α=κcρ=ω(x2x1)22(φ1φ2)log(A1A2). \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.

Was the post useful? Feel free to donate!

DONATE