I’m going to ask my question, explain the problem, show what I have done, and then re-state my question.

(Note, perhaps this is a question better suited for https://stats.stackexchange.com/ or https://math.stackexchange.com, if so, please advise)

Question: How can one model liquid density as a function of pressure and temperature?

Explanation of the problem: The complexity of molecular interactions in liquids precludes development of a general analytical equation of state for liquids analogous to the ideal gas law. For this reason "equations" of state are normally ad hoc formulae based on curves fitted to experimentally measured values of density as functions of temperature and pressure, and possibly chemical factors. One such "equation of state" for liquid can be written as (Furbish, 1997*):

$$\tag{1} \rho = \rho_o [1-\alpha (T-T_o)+\beta (p-p_o)]$$

where the reference density $\rho_o$ is chosen so as to be in the midst of the range of densities likely to occur in the problem of interest. This is to ensure the coefficents $\alpha$ and $\beta$, defined with respect to $\rho_o$, are essentially constants over the associated ranges of $T$ and $p$. For clarification, $\alpha$ denotes a local isobaric coefficient of thermal expansion, defined by

$$\tag{2} \alpha=\frac{1}{V_o}\frac{\partial V}{\partial T}=-\frac{1}{\rho_o}\frac{\partial \rho}{\partial T}$$

and $\beta$ denotes a local isothermal compressibility, defined by

$$\tag{3} \beta=-\frac{1}{V_o}\frac{\partial V}{\partial p}=\frac{1}{\rho_o}\frac{\partial \rho}{\partial p}$$

However, what I would like to do, if possible, is to expand Eqn (1), such that $\alpha$ and $\beta$ are not constants, but rather functions of pressure and temperature, i.e., $\alpha(p,T)$ and $\beta(p,T)$.

What I have done: The following table is density data of a pure liquid (as a function of pressure and temperature) acquired through a constant composition expansion experiment:

enter image description here

The figure below is a snapshot of this density data (as a function of $p$ and $T$): enter image description here

It is clear from looking at the bottom left plot of the figure above that $\beta$ is not a constant, and is in fact a function of pressure and temperature. Likewise, looking at the bottom right plot, the same can be said for $\alpha$.

As a first step, I tried to create a model of this density data in the following manner: Looking at the left plot in the figure below, I first assumed that density increased linearly with pressure (although it is apparent that this is not true), and that this linearity holds for the varying temperatures (again, not true). Then, considering the plot on the right, which is a plot of the slopes, $m(T)$, of the linear trendfit lines relating $\rho$ to $p$ as a function of temperature, a polynomial trendfit was applied to this data of slopes (as a function of $T$), resulting in a 3rd-order polynomial:

$$\tag{4} m=m(T)=a_o+a_1T+a_2T^2+a_3T^3$$

enter image description here

Similarly, a curve can be fit to the data relating $\rho$ to $T$ for some pressure in the midst of the data set as shown in the figure below, here I chose $p=14063$ psia. This curve is approximated by a 3rd-order polynomial:

$$\tag{5} \rho(p,T)=p(14063,T)=b_o+b_1T+b_2T^2+b_3T^3$$

enter image description here

The model for density $\rho$ is then given by combining Eqns (4) and (5):

$$\tag{6} \rho(p,T,p=14063)=(a_o+a_1T+a_2T^2+a_3T^3)(p-14063)+(b_o+b_1T+b_2T^2+b_3T^3)$$

where the notation: $\rho(p,T,p=14063)$ indicates that $p=14063$ psia is the reference pressure used in fitting Eqn (5). Expanding the term in Eqn (6) involving $p$ and rearranging gives:

$$\tag{7} \rho(p,T,p=14063)=(a_o+a_1T+a_2T^2+a_3T^3)p+(A_o+A_1T+A_2T^2+A_3T^3)$$

where $A_i=b_i-14063\ a_i$. Eqn (7) can be abbreviated as:

$$\tag{8} \rho=a_i(T)p+A_i(T)$$

where: enter image description here

The accuracy of this equation to model the experimental data is summarized by the following two plots shown in the figure below. The left plot in the figure is the parity plot showing the quality of fit of the density values estimated using this model, Eqn (7), compared the experimental values. Looking at the plot to the right, the model does a good job estimating density for pressures near 14063 because of the method I used to model the equation. enter image description here

But now, after explaining how I approached this problem (all credit to Furbish, 1997* for explaining this approach) and the assumptions/methods I used, I am hoping to get some feedback on how I might better model this experimental data. As Furbish noted, perhaps the most straightforward approach is to use some sort of multiple-regression technique to fit a function to the data, where the function $\rho(p,T)$ is envisioned as a surface over the $T$ and $p$ coordinates (this is something I am not knowledgeable about and would like to learn to do for this problem). But perhaps there is another way, for example, working with one of the liquid EOSs given on the wiki page.

Restating the question: So, how might one model liquid density as a function of pressure and temperature?

*Furbish, D. J. (1997). Fluid Physics in Geology: An Introduction to Fluid Motions on Earth's Surface and Within Its Crusts. New York, NY: Oxford University Press.

Solution Update 1

Using MS Excel's Solver, I attempted the approach Chester Miller had suggested. First, I performed a least squares minimization fit to the data using the functionality:

$$\tag{9} \ln{\rho}=\ln{\rho(T_0,P_0)}-\alpha(T-T_0)+\beta(P-P_0)$$

resulting in the following 5 parameters to the fit (rounding):

$\rho(T_o,p_o)=0.7851$ g/cm^3

$\alpha=3.24 \times 10^{-4}$ T^-1

$\beta=4.129 \times 10^{-6}$ psia^-1

$T_o=236.65 \ ^\circ$F

$p_o=14038.87$ psia

Using these parameters in Eqn (10)

$$\tag{10} \rho=\rho(T_0,P_0)\exp{[-\alpha(T-T_0)+\beta(P-P_0)]}$$

its accuracy is summarized by the following two plots shown in the figure below. enter image description here

Second, I performed a least squares minimization fit to the data using the functionality:

$$\tag{11} \ln{\rho}=\ln{\rho(T_0,P_0)}-\alpha(T-T_0)+\beta(P-P_0)+\gamma (T-T_0)(P-P_0)+\delta (T-T_0)^2+\epsilon (P-P_0)^2$$

resulting in the following 8 parameters to the fit (rounding):

$\rho(T_o,p_o)=0.8279$ g/cm^3

$\alpha=3.248 \times 10^{-4}$ T^-1

$\beta=5.904 \times 10^{-3}$ psia^-1

$\gamma=2.183 \times 10^{-2}$ (T*psia)^-1

$\delta=1.000 \times 10^{-12}$ T^-2

$\epsilon=9.999 \times 10^{-13}$ psia^-2

$T_o=0.3045\ ^\circ$F

$p_o=26.037$ psia

*note to perform this regression using Excel I had to divide my temperature and pressure values by 1000. The reason being that there are built-in numerical limits in Excel to represent small and large numbers.

Using these parameters in Eqn (11), and using $p$ & $T$ values that have been divided by 1000:

$$\tag{12} \rho=\rho(T_0,P_0)\exp{[-\alpha(T-T_0)+\beta(P-P_0)+\gamma (T-T_0)(P-P_0)+\delta (T-T_0)^2+\epsilon (P-P_0)^2]}$$

its accuracy is summarized by the following two plots shown in the figure below. enter image description here

When performing these regressions with MS Excel's Solver, I constrained the variables $\rho(T_o,p_o)$, $T_o$, and $p_o$ to be values no greater than or less than the maximum and minimum values of pressure, temperature, and density values of the experimental data set.

If you would like to have the excel document that I used to perform these regressions you can email me (see my profile). I am interested in any comments with regards to the goodness of fit of these two models.

  • $\begingroup$ I think that the problem with your final expression for the density $\rho(p,T)$ shown in eqn #7 (or eqn #8) is that you still have a function that is only linear in the pressure $p$ whereas it's quite apparent from your small density vs. pressure plot that assuming a linear $p$ vs. $\rho(p,T)$ dependence won't work very well. I would start off by expanding $\rho(p,T)$ as a Taylor series in $p$ and $T$. So linear terms will involve $p$ and $T$. Then you have quadratic terms $p^2$, $T^2$ and $pT$. Then third-order terms $p^3$, $p^2T$, $pT^2$, and $T^3$, etc.. Proceed until the fit looks good. $\endgroup$ – user93237 Feb 23 '18 at 21:49
  • $\begingroup$ Do you know the chemical composition of the liquid? $\endgroup$ – Nat Feb 24 '18 at 6:43
  • $\begingroup$ @Nat I can find that out for you. In the meantime, what are your thoughts/how might you use the chemical composition to better model the data? $\endgroup$ – Armadillo Feb 24 '18 at 19:20
  • $\begingroup$ To and P0 should have been chosen near the center of the full range of values, as should $\rho(T_0,P_0)$. The regression should have been done on $\ln{\rho}$, rather than $\rho$ itself. The better fit of your original parameterization is puzzling, since the new parameterization basically includes the original parameterization (except for the cubic terms on T). This suggests that the fit to the new parameterization did not represent the global minimum. How do you know that the global minimum was obtained? $\endgroup$ – Chet Miller Feb 25 '18 at 1:02
  • $\begingroup$ @jakemcgregor There're a few ways chemical composition can matter. To just pick a few out: (1) chemical equilibria can shift with temperature/pressure, such that the composition of the liquid becomes different; (2) phase separation can occur (both at macroscopic and microscopic levels); (3) chemical models can be used to better predict the behavior of liquids if you know what species are in 'em. $\endgroup$ – Nat Feb 25 '18 at 3:39

I would do this very differently. Your Eqns. 2 and 3 suggest that a more appropriate form of functionality to start with would be $$\rho=\rho(T_0,P_0)\exp{[-\alpha(T-T_0)+\beta(P-P_0)]}\tag{1}$$or, equivalently, $$\ln{\rho}=\ln{\rho(T_0,P_0)}-\alpha(T-T_0)+\beta(P-P_0)\tag{2}$$ That is, I would work with the natural log of $\rho$ rather than $\rho$ itself. This should also go a long way to more nearly linearizing the behavior with temperature and pressure, and should also help to minimize the percentage errors in density (rather than the absolute errors).

As it stands, my Eqn. 2 represents a 5 parameter fit to the density data ($T_0$, $P_0$, $\rho(T_0,P_0)$, $\alpha$, and $\beta$). I would start out by doing a least squares minimization fit to all the data using this functionality. This would involve reasonable initial guesses for the 5 parameters appropriate, say, to the middle of the range of data.

If the residual errors were found to be too large using the 5 parameters, I would then add a cross-coupling parameter and quadratic terms involving $(T-T_0)^2$ and $(P-P_0)^2$ to the functionality:$$\ln{\rho}=\ln{\rho(T_0,P_0)}-\alpha(T-T_0)+\beta(P-P_0)+\gamma (T-T_0)(P-P_0)+\delta (T-T_0)^2+\epsilon (P-P_0)^2\tag{3}$$

Then, I would re-implement the least squares fit to find a new optimized set.

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  • $\begingroup$ Ah, just to echo what you have said to ensure I understand: Looking at my Eqn 2, $\beta=\frac{1}{\rho_o}\frac{\partial \rho}{\partial p}$, it can be rearranged and integrated to give $$\beta\ \partial p = \frac{\partial \rho}{\partial \rho_o}$$ $$\beta \int_{p_o}^p \partial p = \int_{\rho_o}^{\rho} \frac{\partial \rho}{\partial \rho_o}$$ $$\beta(p-p_o)=\ln(\rho)-\ln(\rho_o)=\ln(\rho/\rho_o)$$ $$\exp{[\beta(p-p_0)]}=\rho/\rho_o$$ $$\rho=\rho_o\exp{[\beta(p-p_0)]}$$ Similar steps can be made for $\alpha$ (my Eqn 3), combine results, and that is how you see to start with your Eqn 1? $\endgroup$ – Armadillo Feb 23 '18 at 23:19
  • $\begingroup$ Yes. That’s right on! $\endgroup$ – Chet Miller Feb 23 '18 at 23:40
  • $\begingroup$ and to gain some more insight, now that I have $\rho(p)=\rho_o\exp{[\beta(p-p_0)]}$ and $\rho(T)=\rho_o\exp{[-\alpha(T-T_0)]}$, how does one decide how to combine the two? I see that $a\exp(X)*b\exp(Y)=ab\exp(X+Y)$, which I suppose is how you arrived to your Eqn 1, but what is the logic/justification for multiplying the two exponentials rather than adding them? $\endgroup$ – Armadillo Feb 24 '18 at 0:55
  • $\begingroup$ Does my equation satisfy the two defitions exactly or not? $\endgroup$ – Chet Miller Feb 24 '18 at 1:31
  • $\begingroup$ Ah! If , for example, $\beta=0$, your eqn results to $\rho(T)=\rho_o\exp{[-\alpha(T-T_0)]}$. However, if one tried to add the two eqns as I had previously asked, one would end up with $\rho=\rho_o(T)\exp{[-\alpha(T-T_0)]}+\rho(p)$ which is incorrect. Thank you, Chester! $\endgroup$ – Armadillo Feb 24 '18 at 1:45

Even for extremely idealized system like a Lennard-Jones fluid there is no analytical equation of state available, and only (very complicated) empirical expressions which are valid in limited pressure and temperature ranges are available (see for example here).

Sure, you can fit $p(T,\rho)$ with a complex enough polynomial or with some other function, but the resulting expression will be valid only in the range of the data from which you obtained the fit. Also, to obtain a good fit you would probably be using a lot of parameters; however, as John von Neumann once said: "With four parameters I can fit an elephant, and with five I can make him wiggle his trunk". Given enough parameters, you can fit any data set to any functional form, but very little physical insight is obtained from such a procedure (in other words, would you be able to give a physical interpretation of all the parameters you are using?).

In conclusion, my suggestion is this: take data from the range of pressures and temperatures you are interested in and find whatever fits them best. You will thus have an empirical formula that works pretty well in that range (and only in that range!).

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