SKH Thermodynamics Labs

Lab 05 · LLE parameter fitting

Does a perfect fit identify the model?

One tie line at 298.15 K was generated by NRTL with τ₁₂ = 3, τ₂₁ = 2 and α = 0.3. Fit τ₁₂ and τ₂₁ with α fixed, and check whether the known values are recovered.

Fit the synthetic control and a measured tie line, then explain what an exact fit shows and what it does not.

2027 prototype
Review the student workflow

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Core: fit, read the fitted model and explain the result. Steps 1–5.

Graduate: the core steps plus temperature transfer, fixed-α identifiability and measurement refits in Step 4.

Graduate route. Complete Steps 1–3 to verify the calculation. In Step 4, test temperature transfer against held-out tie lines, compare fixed α values and refit within measurement bounds. Finish with evidence-backed claims. Go to the graduate investigation.

Choose a data setAll cases fit two τ values at fixed α = 0.3. The measured cases use cyclohexane (1) + methanol (2).

From a tie line to parameters

Two equations, two unknowns
Data · x₁α and x₁β

One measured tie line: the compositions of two coexisting liquids at one T and P.

Equations · μ₁ and μ₂

Equal chemical potential of each component in both liquids gives two equations.

Unknowns · τ₁₂ and τ₂₁

With α fixed, two τ values can usually satisfy both equations exactly.

An exact fit uses all the information in the tie line. Evidence that the model is right has to come from data the fit did not use, such as tie lines at other temperatures.

1. Choose the data and fit settings

Binary · isothermal · one tie line
T / Kx₁αx₁βExpanded U (k = 2)

α = 0.3 is a teaching choice, not a value from the article. Range 0.1–0.6.

The fit also runs six fixed starts and reports how many converge. Search bounds: −2 ≤ τ ≤ 6.

Optional. Your choice does not change the calculation.

Add a fitted result to the comparison before changing a setting.

What is minimized?

For each component i, the residual is ri = ln(xiγi)α − ln(xiγi)β. The objective is the mean of r₁² and r₂². It is unweighted and dimensionless; it is not a maximum-likelihood fit of the measured compositions. A bounded Nelder–Mead simplex stops when its size falls below 10⁻⁸. After fitting, the lab solves the global common tangent of the fitted model and compares the predicted liquids with the data.

Open the current Lab 05 for pasted data, synthetic noise, the objective surface and convergence plots.

2. Read the fitted parameters and check them

Fit the tie line to see the parameters, the predicted liquids and the stability checks.

4. Check your understanding

Write first, then compare

Use the calculations above to answer each question in your own words. Then open the explanation and compare. Your answers are saved with your study.

A. Why does the fit reproduce the tie line exactly?

Fit the measured case at 298.15 K and read the chemical-potential residual and the paper check.

Compare with the explanation

The residual is about 10⁻¹⁰. One tie line gives two equations, equal μ₁ and equal μ₂ in both liquids, and the fit has two unknowns, τ₁₂ and τ₂₁, with α fixed. A two-parameter model can usually satisfy two equations exactly.

So the near-zero residual shows that the tie line was reproduced. It does not test whether NRTL with these τ values describes the mixture at any other condition.

B. Why can constant τ not follow the measured tie lines?

Compare the measured x₁α and x₁β at 278.15 K and 316.15 K with the fitted values at 298.15 K.

Compare with the explanation

With α fixed, gE/RT depends only on composition and the two τ values. If τ is held constant, the predicted coexisting compositions are the same at every temperature: 0.177 and 0.876.

The measured gap narrows from 0.104–0.942 at 278.15 K to 0.301–0.702 at 316.15 K, as the system approaches its upper critical solution temperature. The interaction parameters must decrease as T rises, for example τ = a + b/T.

C. Why is a converged fit not automatically an acceptable one?

Set α = 0.47, fit the measured tie line at 298.15 K, and read the verdict and the Gibbs plot.

Compare with the explanation

At α = 0.47 the optimizer converges, but no τ pair between −2 and 6 makes both chemical potentials equal: the predicted liquids miss the data by about 0.40 in mole fraction.

Converged only means the search stopped moving. An acceptable fit needs equal chemical potentials at the observed compositions and a global common tangent that puts the predicted liquids at those compositions.

4. Investigate what the fit can support

Master's core work

Fit a tie line in Step 1 before running an investigation.

A. Separate calibration from validation

  1. Load the measured case at 298.15 K and fit it. Note the residual and the composition error.
  2. Predict every other reported tie line with constant τ and with τ ∝ 1/T. Read the held-out error and how many predictions fall within the expanded uncertainty.
  3. Choose a second calibration temperature and repeat with τ = a + b/T. Explain why two temperatures can fit the trend but a third is still needed to test it.

B. Test whether α is identified

  1. Fit the same tie line with α = 0.2, 0.3 and 0.47.
  2. Compare τ₁₂, τ₂₁ and the composition error. Note any case where the optimizer converges but the predicted liquids miss the data.
  3. For the measured data, compare the held-out error of the α values that do reproduce the tie line.

C. Refit within measurement bounds

  1. Start with the half-widths shown. For measured data they equal the source's expanded uncertainties; explain why they are useful here and why they are not standard deviations.
  2. Run four one-at-a-time and four corner refits. Compare the τ ranges with the differences between α values in B.
  3. Explain why x₁α and x₁β errors move τ₁₂ and τ₂₁ in different directions.
mol/mol
mol/mol

D. Defend a limited, reproducible conclusion

Submission standard: report the data source and fit settings; cite numerical results from A, B and C; separate calibration from held-out evidence; and propose a discriminating measurement. The record check only confirms that the sections and calculations are present. Scientific reasoning still needs instructor review.

5. Compare and save

  1. Fit and save the synthetic control. Confirm that τ₁₂ = 3 and τ₂₁ = 2 are recovered.
  2. Fit and save the measured tie lines at 298.15 and 313.15 K. Compare their τ values.
  3. Answer questions A–C in Step 4 using the saved comparison.
  4. Download the complete study, reopen it, and reproduce a saved case.
  1. Fit and save the synthetic control. Confirm that τ₁₂ = 3 and τ₂₁ = 2 are recovered.
  2. Fit and save the measured tie lines at 298.15 and 313.15 K. Compare their τ values.
  3. Complete investigations A–C and write the graduate claims.
  4. Download the complete study, reopen it, and reproduce a saved case and its investigation record.

Your fitted cases will appear here after you add them.

Loading the measured data record…

Your settings, comparisons and explanations are kept in this open page. Download one study file to keep the complete record and reopen it later. Fits are recalculated on restore.

Data source

Measured tie lines: Atik and Kerboub, J. Chem. Eng. Data 2008, 53, 1669–1671 (doi:10.1021/je8001688). Numerical values from NIST ThermoML, datasets 6 and 7, with compiler-evaluated expanded uncertainties at 95% confidence. Cyclohexane is component 1; α is the cyclohexane-poor liquid.