Stabilizing Wood: Physics, Chemistry, Materials, Techniques, and Performance: "Just the facts Man"

Next posting - Surface Tension.
What gets in the way of filling the porespace with resin? – Surface Tension

Ok … this is the big one to understand – and will take pictures and one equation.

Any time you have an interface between two different fluids, there is a tension in the interface. The reason for this comes from both thermodynamics (and free-energy based “incompatibilities” between the fluids) and also from force balances between the molecules of the fluids. For now – maybe just think of it as a situation where the physics involved do not like the fluids to be intermixed – and so the forces involved try to minimize the contact area between the two. So the situation looks something like this:
upload_2019-7-31_12-21-37.png


If you think about this drawing, the tension in the interface (“surface tension”) will act to pull that interface back into a “flatter” surface, thus reducing the area of contact between the two fluids. By the way – any gas is a fluid, as is any liquid – so the term “fluid” here is pretty general. The contact could be between two liquids (ex. Oil and water), or a liquid and a gas (water and air, oil and air, acrylic resin and air, etc). A vacuum is not quite a “fluid” – but the same rules giving rise to surface tension arise when a liquid (like oil or acrylic resin) has an interface between it and a vacuum – that interface will have a surface tension present.

Think of the interface kind of like the surface of a balloon: if you inflate the balloon, the pressure inside the balloon goes up, and the surface of the balloon develops a tension in it (you feel that tension as the surface of the balloon becoming “harder”). The interface between two fluids is similar – with the only difference being that the tension of the interface (in units of force per unit length) does not go up as the pressure inside the interface goes up – the surface tension is constant defined only by the materials. I know this might sound a little counter-intuitive, but it is the way it is….

Now – any time that interface between fluids is curved, just like a balloon, there is a pressure difference from one side of the interface to the other (higher pressure being on the concave side of the interface). This pressure difference is described by the “Young-Laplace” equation (different disciplines might have different names for this relationship – but the equation is the same). That equation is:

(pressure difference) = 2 * (surface tension) / Radius

Here, “pressure difference” is the pressure difference across the interface, “surface tension” is the surface tension of the interface (remember, this is defined strictly by the two materials involved), and “radius” is the radius of curvature of the interface.

This is pretty visually intuitive: in the following picture:
upload_2019-7-31_12-22-15.png


The interface has a large radius of curvature, and so the pressure difference across the interface is low. On the other hand, in the picture below:
upload_2019-7-31_12-22-27.png


The radius of curvature of the interface is small – and so the pressure on the left side is much higher.

The other thing to know about this is that the Young-Laplace equation describes an equilibrium (“steady-state”) situation. If the radius of curvature and the pressure drop present do not satisfy the Young-Laplace equation, the situation is physically unstable, and either the radius of curvature will change, or the location of the interface will physically move to find a location / geometry where the equation can be satisfied.

Another thing to know is this: there is something very peculiar about where the interface meets the walls around it, that location indicated by the red circle below:
upload_2019-7-31_12-22-58.png

Blown up to a larger scale, that location is a single point that represents the common meeting of three different materials, “fluid 1”, “fluid 2”, and the solid surface bounding them:
upload_2019-7-31_12-23-13.png


Now … the peculiar thing about this point of contact between three materials is that the thermodynamics (really the free energies of the materials) strictly defines the contact angle. That angle is fixed – and it does not depend at all on the pressure from one side of the interface to the other.

The result of all this is that – for an interface (like the interface of acrylic resin and vacuum on the pores) to be stable – both the curvature relationship of the Young-Laplace equation and the contact angle must be simultaneously satisfied. If they are both satisfied, then the interface will stay right where it is. The only way to change that – like to move the interface deeper into the pore system, is to create a situation where the Young-Laplace equation can not be satisfied.

Re-enter into the picture the “model pore system” I started with (I warned you we would be coming back to this……):
upload_2019-7-31_12-23-30.png


Now, there is something truly unique about those “pinch points” I called pore throats. Lets take a really close up look at the one above indicated by the red dashed circle:
upload_2019-7-31_12-23-43.png



Because of the extreme curvature present at that very small point, it is geometrically possible to continuously satisfy the interface contact angle, even though the radius of curvature of the overall interface changes drastically. Three cases below illustrate this (hopefully).

Low pressure across interface:
upload_2019-7-31_12-23-58.png

Higher pressure across interface:
upload_2019-7-31_12-24-9.png

High pressure across the interface:
upload_2019-7-31_12-24-18.png

As the pressure across the interface goes up, the radius of curvature of the interface goes down (satisfying the Young-Laplace equation), contact angle continues to be satisfied because the small curvature of the wall at the pore throat allows that with very, very small movement of the actual contact point. The interface essentially is “stuck” at that pore throat, even though pressure from the outside is going up.

The interface is said to be “pinned” at the pore throat.

The upshot of this, for us, is that as you increase the pressure on the resin outside of your wood being impregnated, the amount of resin penetration is NOT continuous, or proportional to pressure, as you increase pressure. Rather (this is somewhat simplified), you can increase the pressure, and continue to increase the pressure, and those interfaces remain “pinned” at the pore throats, and you will NOT get any more resin into the wood.

(NOTE: I'm running into a limit on uploads - will need to split this posting and continue with a part 2....)
 
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Surface Tension - Part 2:
The upshot of this, for us, is that as you increase the pressure on the resin outside of your wood being impregnated, the amount of resin penetration is NOT continuous, or proportional to pressure, as you increase pressure. Rather (this is somewhat simplified), you can increase the pressure, and continue to increase the pressure, and those interfaces remain “pinned” at the pore throats, and you will NOT get any more resin into the wood.

That changes, suddenly, when you increase the pressure so much that you get a situation like that depicted below:
upload_2019-7-31_12-27-17.png

The interface has been “pushed through” the pore throat. In this case, the contact angle continues to be satisfied, BUT the Young-Laplace relationship is NOT satisfied (the radius of curvature has gone UP, which should imply a lower pressure across the interface, BUT the pressure difference across the interface has actually gone UP. This is a physically unstable situation …. So the system will change to “find” a stable situation. That situation will be found deeper in the pore system at some smaller pore throat. The interface “jumps” forward (quickly!) and will then settle at some smaller pore deeper in the system (this process is called a “capillary jump”). The people getting oil out of the ground know this well – they use water to “push” oil towards a collection well – and they run in to the same issues regarding pressure and pore sizes in the rock).

One thing to note here is important: the minimum radius of curvature that the interface can take (and thus the pressure it can sustain before it “jumps” deeper into the pore system), is physically defined by the diameter (actually radius) of the pore throat through which you are trying to push that resin. Below that threshold pressure, you get no movement. Above that pressure, the interface jumps forward, and you get to fill more of the pore system (but not all of it – because the next smaller pore will require some higher pressure, etc, etc).

This is quantitative. We know (roughly) the pore sizes of the various types of wood we are trying to fill with resin, and we know the surface tension of the resins – so we can calculate the pressures needed to fill the various pores in the wood. If you want to get fancier, we know (again roughly) the pore size distributions of various woods (covered in an earlier posting) – so technically you can calculate what percentage of the porespace you will fill given a certain wood species and an applied pressure.

I plan on doing that calculation for the next posting – but this is long enough for now, and will probably take a little for you all to digest. There are also some other issues around resin viscosity and time … but will cover that later. So, I think I will stop here for now……
 
RESULT: Pressure needed to "push" resin through pores....

this turned out to be quick to do ... so I will post this quickly and in short form:

Looks like the surface tension of most of the acrylic monomers hover around 25 dynes/cm (0.025 N/m). Not a lot of variation around that value ... and several sources agree - so lets go with that. (if wanted I can post references...)

Also, remember from one of the first postings I wrote: "so - to generalize even more, we have some pores that are about 10 microns, some that are about 1 micron, some that are about 0.1 micron, and some that are 0.01 micron in diameter."

So - churning the numbers (but not showing the calcs) I get the following as pressures needed to "push" the resin through those four classes of "pinch points" (i.e. pore throats):

10 micron pore throat diameter: 1.45 psi
1 micron pore throat diameter: 14.5 psi
0.1 micron pore throat diameter: 145 psi
0.01 micron pore throat diameter: 1450 psi

I will leave that sit for a while, and let you guys digest and talk about this a little, and see what conclusions you reach without my biasing (so far, what I have posted is factual (or should be). The question I think - which is not "fact" but a decision for each of you to reach, is what does that list of pressures versus pore sizes cause you to decide to do??? (but do go back and refer to the earlier post where I gave some examples of pore size distributions versus species of wood......
 
Don't think there's no interest in this subject, sir. I can get deeply involved in the "how" and the "why". This has interested me as much as Larrin's posts on knife steels, science and myths, and toughness tests.

Thank you for your contributions!
 
Given atmospheric pressure is 14.7 psi, those numbers indicate to me that atmospheric pressure alone will get penetration of resin into pores down to 1 micron. On your earlier bar graphs, maple was almost entirely 1 micron or larger, so that might be a reason why maple has generally good results with stabilizing.

The jumps to get down to the .1 and .01 micron pores is pretty large in pressure. Most resin casting pressure pots that I’m aware of max out around 50 psi, so that won’t get oh to those. I somewhat wonder how K&G reaches the claimed 4000 psi. Having a compressor that does 4000 psi isn’t crazy, but it would seem to me it matters the actual psi in the pot not the psi coming out the hose. I wonder what they actually mean when they say they use “4000 pounds”? Truly 4000 psi in some kind of crazy pressure chamber or really just 50 psi that had been filled up with a 4000 psi compressor? If they really have a special chamber not readily available to the home stabilizer that might be your biggest difference right there. For my simple home stabilization I rely only atmospheric pressure alone.
 
Given atmospheric pressure is 14.7 psi, those numbers indicate to me that atmospheric pressure alone will get penetration of resin into pores down to 1 micron. On your earlier bar graphs, maple was almost entirely 1 micron or larger, so that might be a reason why maple has generally good results with stabilizing.

The jumps to get down to the .1 and .01 micron pores is pretty large in pressure. Most resin casting pressure pots that I’m aware of max out around 50 psi, so that won’t get oh to those. I somewhat wonder how K&G reaches the claimed 4000 psi. Having a compressor that does 4000 psi isn’t crazy, but it would seem to me it matters the actual psi in the pot not the psi coming out the hose. I wonder what they actually mean when they say they use “4000 pounds”? Truly 4000 psi in some kind of crazy pressure chamber or really just 50 psi that had been filled up with a 4000 psi compressor? If they really have a special chamber not readily available to the home stabilizer that might be your biggest difference right there. For my simple home stabilization I rely only atmospheric pressure alone.

Thanks - that is the kind of thought/discussion I was hoping people would pick up. You ask an interesting question - and though I dont have the direct answers re. pressure - I think you are getting there. One example I can give is SCUBA tanks. the standard aluminum SCUBA tank is designed to hold 3000 psi. look them up on the web - they are small enough to carry on your back, and even then they are HEAVY - with very thick walls and only a single threaded opening at the top. Certainly then, there are compressors capable of hitting those pressures. But A larger container with an opening large enough to put wood blocks into and out of, and which could hold those pressures would, in your words, be "some kind of crazy pressure chamber". Certainly doable -- but most definitely way outside of what might even be considered at home. I truly had not even been thinking along those lines - but I think you are on to a reasonable train of thought...

I would take K&G at their word ... which means for the really dense woods with the majority of pores below 0.1 micron, home treatment does not stand a chance of penetrating most of the pores.

The kind of question I started out asking is .... what if 50% of the pores were 1 micron or larger. Would home treatment be "good enough"???
 
You are forgetting about something.

Some pores already have stuff in them.

Heartwood is dead wood, and what makes heartwood different from sapwood is the presence of heartwood extracts. Its something of a catch all term, but there are a wide variety of aromatics, oils, resins, lignins and tannins that both saturate the wood and become deposited into the pores of some woods. These both block the physical travel of resin through the pores, as well as interfere with curing, as they presence of all these organic compounds can prevent a proper cure.
 
You are forgetting about something.

Some pores already have stuff in them.

Heartwood is dead wood, and what makes heartwood different from sapwood is the presence of heartwood extracts. Its something of a catch all term, but there are a wide variety of aromatics, oils, resins, lignins and tannins that both saturate the wood and become deposited into the pores of some woods. These both block the physical travel of resin through the pores, as well as interfere with curing, as they presence of all these organic compounds can prevent a proper cure.
Thanks Ben. Yeah - when i am thinking/ talking about pores, i am talking about small open spaces where it is potentially feasible to fill those spaces with resin. If they are already filled with extracts from the wood, from that perspective they cease to be “pores” (they would be invisible to mercury Porosimetry).

A question though: if these are already “filled”, is the same purpose as filling them with resin achieved (at least somewhat), and so do you not need to worry about them from that perspective?

Bigger worry is the potential chemical interference with resin curing - i was not aware of that. Do you know what K&G does to mitigate that ..? . or is it something you just need to live with with and be aware of, wheher you are K&G or a home stabilizer?
 
My understanding is those woods with high natural oils and resins don’t take well to stabilizing regardless of home or professional. My understanding is also those woods generally don’t actually need stabilizing because they are naturally stable. Wood has been used for a long time before stabilizing was a thing by selecting woods known to do well by basic human trial and error.

This raises another question in my head. The purpose of stabilizing is to fill the pores with resin so that water cannot enter the pores to cause expansion (or conversely to have been in the pores and leave them to cause shrinking). Given water will never be forced into the wood at higher than atmospheric pressure, is the use of pressurization to force resin into pores that cannot be filled by a fluid at atmospheric pressure really necessary. Something like ebony from the chart that is almost entirely needing over 1000 psi of pressure to get a liquid into its pores should not be at any risk of taking on significant water from atmospheric pressure. In the “what’s good enough?” question, isn’t good enough where you prevent moisture entry at atmospheric pressure?
 
The purpose of stabilizing is to fill the pores with resin so that water cannot enter the pores to cause expansion (or conversely to have been in the pores and leave them to cause shrinking).
I believe Ben has written in this thread that stabilizing does not make the wood waterproof. I would think that it might slow down the moisture adsorption / desorption processes - but not stop them. I would also think that the structural reinforcement of the resin might mitigate the degree of swelling/warping, even with moisture penetration - but not totally eliminate it. I am with you, though, on your "whats good enough" question. If what you are after is truly just "stabilization" (structural) - is the penetration/reinforcement you can get with atmospheric pressure enough to make the wood stable enough for your purposes? If, as others have said, you are mostly after weight and "feel" ... that is a different question and different objective....
 
I suppose structural stability might actually be the more accurate objective of stabilizing vs my above comment about moisture uptake (though I think for the purposes of knife makers, moisture uptake is a component of their concern). The comment about structural stability made me think back to where do us knife folks that do play with home stabilizing go for our kit? We buy them from woodworking places. Home stabilizing is a very common practice amongst wood turners. Given they are known to intentionally turn green wood, their purposes would seem less about moisture and more about structural stability. Pieces of burl or spalted wood that would crumble or explode on the lathe need stabilizing to be structurally sound for turning. Again though, it is very common to home stabilize in the woodworking world, and they are going to spin that piece of wood at high rpm in front of their face and stab a sharp object into it! So home stabilizing can make a piece of burl etc. structurally stable enough for that abuse but not structurally stable enough for a knife handle? Then I think about the various “super steels” and contemplate the possibility that for some knife makers the terms “good enough” and “overkill” are actually listed as synonymous in their personal dictionary :p
 
As for chemical interference, there is just sort of a lucky fact that most woods that need stabilizing can take it.

Woods like Desert ironwood, rosewoods, ipe and teak have very minimal weight gain and can sometimes weep resins.

The only woods that would really need and wont take resin are woods like camphor burl
 
So here’s one to ponder for a possible reason. The info from the wood turning folks is for most woods the after vacuum soak time should be double the time under vacuum. With walnut this isn’t a sufficient amount of time, but if the walnut is left to soak for longer (a week recommended) it works. At first I was thinking walnut would be smaller pores, but the logic here is smaller pores would require more pressure. Letting it soak longer doesn’t change the pressure, just the amount of time it’s “under pressure.” Is there something about walnut that would have a logical explanation of not working when left for the standard duration but working if left longer? If that could be identified, the home stabilizer could potentially categorize woods into those that needed shorter or longer soaks.
 
Possibly. If you remember going back to the beginning, i included viscosity and time as things to consider. Now.. i have not done the calculations, but the basic idea is you have fluid trying to flow through a tube, the more viscous the fluid, the slower it flows, you can increase how fast it flows by increasing the pressure “pushing” it. Even if you have a pressure high enough to overcome the surface tension at the pore throats, it is still possible you have a long wait at one atmosphere to “push” the resin 2”or so through the capillaries. Not sure if a week is what we are talking about... but i will run the numbers soon and include in next post.

Great observation - you are anticipating my next points!
 
Possibly. If you remember going back to the beginning, i included viscosity and time as things to consider. Now.. i have not done the calculations, but the basic idea is you have fluid trying to flow through a tube, the more viscous the fluid, the slower it flows, you can increase how fast it flows by increasing the pressure “pushing” it. Even if you have a pressure high enough to overcome the surface tension at the pore throats, it is still possible you have a long wait at one atmosphere to “push” the resin 2”or so through the capillaries. Not sure if a week is what we are talking about... but i will run the numbers soon and include in next post.

Great observation - you are anticipating my next points!
 
What about shaking the chamber?

It sounds funny but I find when I’m under full vacuum (28.9 inches in my set up), that I can shake loose a few pockets of air by smacking the chamber (surface tension being broken?).
I also notice seems to help the wood absorb the resin a little faster.
 
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