4
$\begingroup$

Suppose that I have 10 kg of plastic/polymer that is non solvable in water. I then divide the plastic/polymer into two equal parts. Let us name the parts A and B in order to identify them. I then proceed to make beads from each part. I make $n_A$ beads of radius $r_A$ from part A. I also make $n_B$ beads of radius $r_B$ from part B. Here $r_A>r_B$ and consequently $n_B>n_A$. Suppose that I put the beads from part A into 10 liters of water and I do the same for the beads from part B. Which one is more viscous, the "solution" from part A or from part B? Note: here $n_A,n_B$ stand for some arbitrary number.

$\endgroup$
  • 2
    $\begingroup$ I edited your original question for clarity. In doing so I removed one statement about "dissolving" the beads. Since you state that the beads are insoluble, I think this is OK, but please check to make sure I did not change the meaning of your question. $\endgroup$ – bobthechemist Dec 30 '13 at 13:34
  • $\begingroup$ If the polymer is lighter than water then it will form a scum on the top. If denser then it form a layer on the bottom. Though both depend on the size of beads being big enough to not be affected by the water jostling so it can settle. $\endgroup$ – user2617804 Dec 31 '13 at 7:21
  • $\begingroup$ This is a very interesting question, but I think it would get much more attention, and much better answers, at physics.se. I've tried my best at answering anyway $\endgroup$ – Michiel Mar 24 '14 at 20:31
  • $\begingroup$ @Michiel Thanks for your suggestion, but questions older than 60 days can no longer be migrated. $\endgroup$ – jonsca Mar 24 '14 at 23:04
3
$\begingroup$

Disclaimer: I will be talking about the shear viscosity. I think that most of what I write is also applicable to extensional viscosity, although with different numerical constants, but I am not fully sure so I will limit myself to shear viscosity.

Most theoretical models predict that the shear viscosity of both solutions will be equal. The reason is that these models describe the viscosity of the suspension $\mu$ as the viscosity of the 'empty' liquid $\mu_0$ $+$ a correction term that depends on the volume concentration of solids, $\phi$.

The most famous one is the Einstein viscosity equation that simply reads: $$\frac{\mu}{\mu_0}=1+2.5\phi $$

This model works for low solid concentrations when inter-particle interactions can be neglected. Later Batchelor showed that including long-range hydrodynamic interactions an additional correction of $+6.2\phi^2$ should be included, but still only a dependence on the volume concentration is there.

A nice summary of more models can be found in this paper: K.D Danov, J. Colloid Interface Sci., 235, 144–149 (2001). That paper also includes a discussion that shows that the factor 2.5 should be dependent on the mobility of the particle. This would mean that it will be higher for smaller particles, so in your case that the solution with part B will be somewhat more viscosity.

Another overview of models, including models which have an increasing viscosity with increasing particle size can be found here. So that would point to part A giving the more viscous solution.

So to summarize: to first order (i.e. low concentrations) the two viscosities will be the same. At higher orders you have competing effects of the particle radius that should increase the viscosity and the particle number that should also increase it. Which effect is dominant is probably hard to tell theoretically.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.