Calculating the coefficient of thermal expansion from density

[mark_bose]

TL;DR Summary

Derivation of the coefficient of thermal expansion from density correlation and discrepancy with other experimental correlations.

Hi all,

I am trying to derive a relation (as function of temperature) for the coefficient of linear thermal expansion (CLTE) strating from a correlation for the density. But I'm getting a huge discrepancy from the experimental data.

I'll start telling you some reasoning I did to get a relation for the CLTE:

Let's say that $\alpha(T)$ is the linear coefficient of thermal expansion. Than, the generic length in one dimension should vary as
$$L=L_0(1+\alpha(T)\times (T−T_0))$$
Let's now consider a cube of volume $V_0=L_0^3$. Accordingly to the previous equation, and supposing a isotropic expansion, the new volume should be:
$$V=L^3=L_0^3(1+α(T)\times(T−T_0))^3=V_0(1+\alpha(T)\times (T−T_0))^3$$
Since mass is conserved, one can derive the density ratio $\rho(T_0)/\rho(T)=V/V_0$. Therefore, if I'm not mistaken, the relation for the CLTE is straightforward:
$$\alpha(T)=\frac{(\rho(T_0)/\rho(T))^{1/3}−1}{T−T_0}$$
Now, I tried to apply this approach to the specific case of Uranium-Molybdenum metallic alloy. The paper from Idaho National Laboratory (Thermophysical Propertiesof U-10Mo Alloy D. E. BurkesG. S. MickumD. M. Wachs) provides a correlation for both the density and the CLTE. The problem is shown in the figure below: if i use the density correlation provided by INL ($\rho(T)=17.15+8.63\times 10^{−4}(T[°C]+20)$) and I derive the CLTE from it (blue line), it is different from their correlation (orange line) for CLTE ($\alpha(T)=11.2+8.07\times 10^{−3}T$)

I suspect I made some naive mistake in the derivation of CLTE correlation or I'm looking at two different definition of the CLTE. Any idea?

 

[erobz]

Do you know you are using a first order Taylor Series approximation of the solution for the ODE (1):

$$ \alpha(T) dT = \frac{dL}{L} \tag{Eq.1}$$

?

 

[mark_bose]

Do you know you are using a first order Taylor Series approximation of the solution for the ODE (1):

$$ \alpha(T) dT = \frac{dL}{L} \tag{Eq.1}$$

?

Thanks for your answer. You are right, it is a first order approximation, but I do not think that this is the problem: If so, the first order approximation and the INL correlation should at least agree at low $\Delta T$ but it is not the case. However, to be sure I can derive the real relation.

Let's start from your equation and integrate it in the range $(T_0,T_1)$:
$$ \int_{L_0}^{L_{1}} \frac{dL}{L} = \int_{T_0}^{T_1} \alpha(T)dT $$
It becomes:
$$ ln\left (\frac{L_1}{L_0} \right) = \bar \alpha (T_1 - T_0) $$
Where $\bar \alpha$ is the average value in the aforementioned range. Let us multiply by 3 both side and exploit the logarithm rule: $a ln(x) = ln(x^a)$. It follows:
$$ ln\left ( \left( \frac{L_1}{L_0} \right)^3 \right) = ln \left( \frac{V_1}{V_0}\right) = 3 \bar \alpha (T_1 - T_0) $$
Finally we can introduce the density ratio $\rho_1/\rho_0 = V_0/V_1$ and derive a relation for the average CLTE:
$$ \bar \alpha = \frac{ln(\rho_0/\rho_1)}{3(T_1-T_0)}$$
You can see from the figure that the first order relation and the complete one agree quite well at low temperature (as it should be). However, both are different from the INL correlation.

Any other idea?

 

[erobz]

Is $\alpha$ a function of temperature $T$ or not? Treating it as a constant $\bar{\alpha}$, if its a function of temperature $T$ seems hand wavy.

 

[Lord Jestocost]

TL;DR Summary: Derivation of the coefficient of thermal expansion from density correlation and discrepancy with other experimental correlations.

...Now, I tried to apply this approach to the specific case of Uranium-Molybdenum metallic alloy. The paper from Idaho National Laboratory (Thermophysical Propertiesof U-10Mo Alloy D. E. BurkesG. S. MickumD. M. Wachs) provides a correlation for both the density and the CLTE. The problem is shown in the figure below: if i use the density correlation provided by INL ($\rho(T)=17.15+8.63\times 10^{−4}(T[°C]+20)$) and I derive the CLTE from it (blue line), it is different from their correlation (orange line) for CLTE ($\alpha(T)=11.2+8.07\times 10^{−3}T$)......

$\rho(T)=17.15-8.63\times 10^{−4}(T[°C]+20)$ !!

 

[mark_bose]

Is $\alpha$ a function of temperature $T$ or not? Treating it as a constant $\bar{\alpha}$, if its a function of temperature $T$ seems hand wavy.

$\bar \alpha$ is function of the integration extremes $T_1$ and $T_0$. While $T_0$ is usually taken as environmental temperature, $T_1$ can be arbitrary selected. The meaning of $\bar \alpha$ is the "average CLTE" in the range $(T_0,T_1)$, its value varies if the extremes vary as well. In the graph i wrote T, while probably I should have wrote $T_1$. If you look at the reference from INL, they printed the avg. CLTE as function of the extreme $T_1$ keeping constant $T_0 = 20 °C$. I did in my graphs.

 

[erobz]

$\bar \alpha$ is function of the integration extremes $T_1$ and $T_0$. While $T_0$ is usually taken as environmental temperature, $T_1$ can be arbitrary selected. The meaning of $\bar \alpha$ is the "average CLTE" in the range $(T_0,T_1)$, its value varies if the extremes vary as well. In the graph i wrote T, while probably I should have wrote $T_1$. If you look at the reference from INL, they printed the avg. CLTE as function of the extreme $T_1$ keeping constant $T_0 = 20 °C$. I did in my graphs.

View attachment 355164

But if the average $\bar{\alpha}$ is increasing over sub ranges of $T$, then $\alpha$ ( an apparently increasing function in $T$) cannot be extracted from the integral involving $\text{Eq.1}$, as though it were actually constant over the full range of $T$.

 

[mark_bose]

Isn't it the mean value theroem (Wikipedia)?

By definition, the average value of a function $f(x)$ in the range $(a,b)$ is $\bar f = \frac{1}{b-a}\int_a^bf(x)dx$, consequently:
$$\int_a^b f(x)dx = \bar f (b-a)$$

 

[erobz]

Isn't it the mean value theroem (Wikipedia)?

By definition, the average value of a function $f(x)$ in the range $(a,b)$ is $\bar f = \frac{1}{b-a}\int_a^bf(x)dx$, consequently:
$$\int_a^b f(x)dx = \bar f (b-a)$$

I see what you are saying because of the linearity of $\alpha$ you can skirt the integration.

 

[erobz]

Have you noticed the sign error @Lord Jestocost brought up in post 5?