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Topic: Solvent-Polymer solution drying  (Read 2206 times)

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Offline arnau.castillo

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Solvent-Polymer solution drying
« on: June 30, 2015, 03:16:18 AM »
Hello everyone,
once a Solvent-polymer solution is completely mixed, both in liquid phase, I understand that the drying process is as follows:

1- Solvent starts to evaporate from the surface of the solution.
2- At the same time that solvent is removed from the surface, by diffusion solvent moves up from the bulk of the solution to the surface.
3- This process happens continuously up to polymer solidification (how to define that exactly?)
4- Once solidified, the gradient of concentration pushes again solid diffusion continued with mild evaporation.
5- the process runs infinite, leaving sometimes solvent in the final polymer (solvent retention).

My questions are: Is that the correct approach to the topic? Are my steps correct? How can I model and calculate each step?  Thanks to everyone for your help beforehand!

Offline Enthalpy

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Re: Solvent-Polymer solution drying
« Reply #1 on: June 30, 2015, 05:21:28 AM »
¡Hola Arnau!

It is my understanding too.

Then, to compute... You need figures hard to find:
- The ease of solution of the solvent in the solid polymer
- The diffusion coefficient of the solvent in the solid polymer
- or rather a combination of both: the diffusivity
so you may have to determine them experimentally by yourself. I have very little data, only for gases in solid polymers.

Modelling the process with letters instead of the real data is easy (but brings little)... This is a diffusion equation.
- Neglect the early duration it takes to evaporate the solvent when the polymer is still dissolved or soaked, because then, the polymer isn't tight, so this goes quickly.
- Assume that the diffusion constant doesn't depend on the solvent concentration in the already solid polymer. This can be badly wrong; I expect it to hold only at small concentration.
- Then the solvent flow per surface unit at some depth is proportional to the concentration gradient, and
- The local gradient of the flow tells how quickly the concentration varies there.

That is, you must get da/dt = d2a/dz2 where a is the concentration and z the depth, and I didn't write the diffusivity constant.
https://en.wikipedia.org/wiki/Diffusion_equation

Then, you may suppose that
- The initial distribution was flat (after some time, it doesn't matter much)
- The solvent disappears so efficiently from the surface that its concentration is zero there
- The substrate holding the polymer is tight to the solvent (if not, you need data for the substrate as well: a first varnish layer dries quicker on wood as the solvent gets absorbed)

Under these assumptions, you model the layer on a reflecting substrate by a layer double as thick with both surfaces exposed to air. An algebraic solution is known as well: it's (after a short time) a sine of depth with nodes at both surfaces and it decreases as exp (-t/τ) where τ is proportional to the squared thickness.

This is exactly Fourier's theory, for which he developed his Fourier series - he did it for heat instead of solvent concentration. The higher harmonics of the initial solvent distribution vanish quickly (32, 52... faster), leaving only the longest-range sine.
https://en.wikipedia.org/wiki/Fourier_series

When the solvent concentration is still high, evaporation may limit the speed. You should get again an exp (-t/τ2) with a faster τ2 that depends on the non-squared thickness.

For both processes, you might model the temperature effect by activation energies: exp(-E/kT).

Err... My approach would be to
- take the exp (-t/τ) solution
- put a layer of dissolved polymer on very sensitive scales or under a very sensitive solvent sniffer. Know the final thickness from the mass.
- get an experimental curve of the solvent amount remaining in the polymer.

With limited luck, the curve fits the exp (-t/τ), and then you deduce the materials' diffusivity and can make predictions with reasonable confidence.

Offline arnau.castillo

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Re: Solvent-Polymer solution drying
« Reply #2 on: July 06, 2015, 09:48:32 AM »
Hello Enthalpy,
first of all, thanks for the long and detailed email! ;-)

Might be because I have no chemistry background that I'm missing some parts or, not understanding it properly.

1- Both evaporation and diffusion processes are rulled by Fick's diffusion equation? I thought that the Polymer-Solvent area was ruled by Fick difussion (both liquid and solid phases) and the solvent-air area was ruled by the Chapman–Enskog theory ([https://en.wikipedia.org/wiki/Mass_diffusivity]), as it contains evaporation rate and flow effects in one. Is that correct?

2- For the diffusion part, I understand that I can use the Arrhenius equation with given Activation energy (Qd) and Do, to estimate D through all the ranges of Temperatures (D=Do*exp(-Qd/RT)). Then, Considering D=exp (-t/τ2) constant, I can find the concentration to any point. But I have a doubt: If my room temp is constant, does this Temp in the Arrhenius equation changes? is this T the solvent polymer reaction temp? How do I calculate that?

3- Ok, sumarizing and drawing again the conceptual problem: I have a solvent-polymer mixture at flat concentration first. This liquid mixture runs with the Stokes–Einstein equation for liquid diffusion. Until when? Would be a good assumption to assume is solid once viscosity is below 5% of the expected final viscosity? And later, once solid, and considering concentration at the boundary with the air 0, we can apply the mentioned diffusion equation. While all of this happens, evaporation + air diffusion happened with the Chapman constant. How to consider the concentration at surface? Always flat?

I understand the different processes, but it is complicated to define how to interconnect them. Thanks to everyone and excuse the conceptual mess!! ;-)

Offline Enthalpy

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Re: Solvent-Polymer solution drying
« Reply #3 on: July 08, 2015, 05:45:27 AM »
1 : I supposed (I may be wrong) that the slowest process is the diffusion of the little remaining solvent in the solid polymer, and that the evaporation of the solvent is much quicker, hence I had neglected the evaporation.

2 : If the temperature stays constant, then you don't need Arrhenius nor the corresponding activation energy. Simpler. Arrhenius would only be a way to measure a speed at a few temperatures and interpolate the speed to other temperatures.

3 : τ2 corresponds to the slowness of evaporation in case it is important (I supposed no in 1).

Experimentally, you might proceed that way:
  • Have a spot of dissolved polymer on sensitive scales. From the mass, area, density you know the thickness.
  • Measure the mass over time. At the end (or from the manufacturing method) you know how much solvent and how much polymer there was. Do it with air humidity and wind speed equal to the targeted use, or using artificial but easily repeatable conditions.
  • Draw a curve of the logarithm of the remaining solvent amount over the linear time. With some luck, you observe two epochs, each showing a straight line, the first epoch decreasing faster.
  • Attribute the first, faster mass loss to evaporation speed and identify τ2, and the second, slower mass loss to solvent diffusion in the solid polymer, and identify τ there.
  • If any necessary, repeat at several temperatures to identify activation energies E2 and E. This may be needed just because both processes are so sensitive to temperature.
  • If the two epochs and lines aren't so distinct, try to model a sum of both slownesses or some sort of combination...
  • The evaporation and diffusion models predict a dependency over the thickness. A few measures more let you check if the interpolation works.
  • Different thicknesses are also a way to distinguish τ2 from τ if the initial thickness doesn't show two clear epochs.

Varnishes for instance grow a thin solid membrane at the exposed surface while the depth is still liquid, less easy to model. If this epoch lasts for long, it needs its own model. A perfectly dry polymer layer over a solution would be easy to model (skin thickness increasing as sqrt(t)) but not necessarily accurate.

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