I know a larger part of this question is "it depends", since scales matter here. Let's bound ourselves to desktop 3D-printed objects, which will typically be no more than a few cm thick.

After an hour or so of acetone vapor smoothing, some people have noticed that ABS parts can bubble a week or two later. Below is an example of where an ABS part deformed when exposed to constant heat.

enter image description here

It makes sense to me that once the acetone is subsurface that it will just as easily continue to work its way deeper into the part as head toward the surface and evaporate.

On a long enough time line, though, all the acetone will eventually work its way to the surface and liberate itself from the plastic.

How can I determine that timeline? I would have thought it was hours, but empirical evidence seems to suggest it's at least weeks, if not longer.

(I'm making an assumption here and that's that acetone is a harmless solvent and will not permanently modify the plastic. It's unclear if that assumption is completely correct.)

  • $\begingroup$ If you're going to downvote, esp. if you're downvoting a new contributor, please take the time to explain why you think this question is not appropriate for SE.Chemistry. $\endgroup$ Oct 17, 2022 at 22:18
  • $\begingroup$ Since acetone turns a solid object into goop, why do you consider it “harmless”? Do printed objects not exposed to acetone behave the same way? $\endgroup$
    – Jon Custer
    Oct 18, 2022 at 14:10
  • $\begingroup$ Since, as you state, the effect varies, you'd need to experiment with you're printer and feed stock, to eliminate some of the many variables such as feed composition, printing temperature, heating time, thickness, temperature of piece and acetone vapor in smoothing, temperature and airflow for removing absorbed vapors, etc. $\endgroup$ Oct 19, 2022 at 0:03
  • $\begingroup$ @JonCusterr "harmless" in reference to the material chemistry. I.e., water will dissolve table salt, but when the water evaporates the original salt remains chemically unchanged. I don't know if this is true about acetone and ABS. $\endgroup$ Oct 19, 2022 at 9:19
  • 1
    $\begingroup$ @KennSebesta, rigorously, the solvent never completely leaves the workpiece; it's depletion curve is logarithmic, and, though concentration decreases over time, consider that it halves, and halves again, and halves yet again. It also would not be uniform with depth. If you could measure the rate, and establish what would be an allowable remainder of solvent, you could find this yourself. $\endgroup$ Oct 19, 2022 at 14:26


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