I am trying to calculate the density of a magnesium sulphate solution and came across this paper which determines an equation from experiment to do that. However, I am having trouble understanding the paper, as parameters A, B, C and D are mentioned but I don't understand how to calculate these from Table 2 in the paper. Can someone please explain how this works?


I am adding the data from the above paper: enter image description here

The paper mentions the equation enter image description here

to calculate the relative density where the parameters A, B, C and D are said to come from the table. m stands for the molality (I assume mol/kg) t is temperature (°C) and d is in g/cm^3. As you can see there is no mention of A, B, C or D in the table, leading to the confusion.

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    $\begingroup$ Unfortunately behind the paywall... $\endgroup$ – Poutnik Nov 15 '20 at 9:24
  • $\begingroup$ Sorry I didn't notice but I still seem to have some kind of access from my university. But there are ways around the paywall $\endgroup$ – hanai Nov 15 '20 at 9:31
  • $\begingroup$ I did not get a clear answer on the SE policies when googling but I'd also prefer not to post potentially illegal links or copyrighted material. $\endgroup$ – hanai Nov 15 '20 at 9:39
  • $\begingroup$ Note that it seems there is an error in the exponent in the formula ( 2/3 versus 3/2 ). Rather check it within the paper context and/or the calculated values. $\endgroup$ – Poutnik Nov 15 '20 at 12:33
  • $\begingroup$ Mistake from my side. corrected $\endgroup$ – hanai Nov 15 '20 at 16:05

Each of parameters A-D is a polynomial function of t:

$$A = \text{"m"} + \text{"mt"}.t + \text{"mt2"}.t^2 + \text{"mt3"}.t^3 + \text{" mt4"}.t^4$$

$$B = \text{"m3/2“} + \text{"m3/2t"}.t + \text{"m3/2t2"}.t^2$$

$$C = \text{"m2“} + \text{"m2t"}.t$$

$$D= \text{"m5/2"}$$

The terms in quote means the equally denoted numerical values from the provided table.

The complete final function $1000 \cdot \Delta \rho=f(m,t)$ has for each term $a_{b,c} \cdot m^a \cdot t^b$ the value for $a_{b,c}$ equal to the particular table value named "$m^bt^c$".

So one can the complicated function use in 2 ways:

  1. The full blown function $1000 \cdot \Delta \rho=f(m,t) = \sum {\left( a_{b,c} \cdot m^a \cdot t^b \right)}$ without the need of coefficients A-D.
  2. Temperature specific $1000 \cdot \Delta \rho=A \cdot m + B \cdot m^{3/2} + C \cdot m^2 + D \cdot m^{5/2}$ with precalculated A-D values for given temperature.

BTW, it looks like a very interesting numerical approximation. Maybe even better could be rational functions like Padé approximant, that have smaller deviations and better extrapolation behaviour, compared to polynomials.

Feedback for the Padé approximation:

For approximating of analytical functions, see e.g Math SE site - how-to-compute-the-pade-approximation. For approximation of experimental discrete data, the best approach is probably numerical mathematics, the least square method combined with function minimum search algorithm. They probably used the same technique for their numerical model with half-integer powers.

Using half-integer power model is unusual. It may work better than polynomials in many cases, but just for positive numbers. Fortunately, they do not calculate with negative Celsius temperature :-) ) Note that Math SE site has even tag pade-approximation . Also, it is included in MATLAB libraries, if applies. See Octave as both online/offline free functional clone of MATLAB.

See also my answer in this Q: Is there a quick way to approximate a logarithm in Nernst equation?.

  • $\begingroup$ Interesting. Do you have some resources on how to use Padé approximation for curve fitting? Aside from that, what is your guess on how they arrived at this kind of equation? The paper doesn't touch on this. $\endgroup$ – hanai Nov 15 '20 at 16:07
  • $\begingroup$ The answer was updated by Padé appr. feedback. $\endgroup$ – Poutnik Nov 15 '20 at 17:43

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