## Bubble Gone!

### Quick Start

If you apply some adhesive (or thermal conductive paste) and squeeze it under glass to see if it makes a nice even layer you often see bubbles form then disappear. Where do they go and why do they go so suddenly?

### Credits

This is part of my mission to emphasise that how you apply adhesive can be as important as what adhesive you choose.

### Bubble Gone

_{o}μm

^{-9}m²/s

^{-5}mol/(m³Pa)

_{Gone}s

```
window.onload = function () {
//restoreDefaultValues(); //Un-comment this if you want to start with defaults
Main();
};
function Main() {
saveSettings();
const inputs = {
Do: sliders.SlideDo.value / 1e6, //μm to m
D: sliders.SlideD.value/1e9, //m units
H: sliders.SlideH.value/1e5, //mol, m, Pa units
P: sliders.SlideP.value, //stay in bar, it doesn't matter
}
//Send inputs off to CalcIt where the names are instantly available
//Get all the resonses as an object, result
const result = CalcIt(inputs)
//Set all the text box outputs
document.getElementById('tGone').value = result.tGone
}
//Here's the app calculation
//The inputs are just the names provided - their order in the curly brackets is unimportant!
//By convention the input values are provided with the correct units within Main
function CalcIt({ Do, D, H, P }) {
const R=8.314 //Gas constant
const T=298 //25C
const t=1/(8*D*H*R*T)*Do*Do*Math.pow(1+P,0.3333)/P
return {
tGone: t.toPrecision(3),
};
}
```

### Disappearing Bubbles

When you apply an adhesive there's a chance you will have trapped an air bubble. If that bubble stays there then you have a focus for crack failure, so you really want it to disappear. The same applies when you use a thermal conductive paste - any bubble defect will be a local hotspot. Fortunately we often squeeze the adhesive or paste (see theStefan's Drop Squeeze app) and it turns out that bubbles can be made to dissolve into and diffuse through the adhesive, with the air probably diffusing slowly out of the joint over time (rather than exploding back when the pressure is removed). As the image tries to show, the diffusion of the gas into the bulk means that the bubble shrinks, though the first reduction in size might have come from applying the squeeze pressure which makes the bubble volume smaller.

If you had an adhesive that was carefully de-gassed and you accidentally created a 400μm bubble, you would typically find it disappearing in a few 10s of seconds. The diffusion coefficient and concentration gradient are both large enough for the air in the bubble to diffuse into the bulk in that time. The radius of the bubble creates a slight excess pressure (from the usual curvature equation `P=(2σ)/R`), and this pressure gets larger as the bubble gets smaller, but even without this pressure, the process is fairly fast.

But our pots of adhesive are generally saturated with air and if you did the same measurement you would find timescales in the order of 1 hour for the same-sized bubble to dissolve away. Fortunately, because we have to squeeze our adhesives, and, because of Stefan, we have to apply a high pressure to really squeeze down to a thin layer, we get a double benefit.

- If you apply a pressure of 2 bar, the diameter of the bubble automatically decreases by a factor of ~1.4 just from simple bubble pressure effects - the first reduction in bubble size in the image.
- The internal gas pressure increases so, by Henry's law, the solubility in the surrounding layer of adhesive increases, so the rate of diffusion increases and the shrinkage continues as in the second reduction in size.

The combination of these two effects is powerful. A bubble that would take 1hr to dissolve with just the slight excess pressure from bubble curvature now takes just a few seconds. That's why our adhesive joints are often better than we deserve - the trapped air just dissolves into the bulk.

### But what about ...

**When the pressure is removed?**Because the gas has dissolved in the bulk, there's no strong driving force to make it re-appear as a bubble. In any case, we know that the pressure of a nascent bubble, with a very small R, is very high, so it is hard for the bubble to re-nucleate.**The viscosity of the adhesive?**Surprisingly it doesn't make all that much difference in many of the simulations because dissolution is a slower process than movement of the adhesive towards the shrinking bubble. Maybe viscosity slows down bubble collapse in the few second range, but we probably don't worry too much about that. A full-scale simulation*can*be written to include the effects, but I've chosen a formula that captures very well lots of the experimental evidence for polymers in the 3000 Pa.s viscosity range, so the times here may be somewhat pessimistic for low viscosity adhesives. However, if we have higher viscosity we will need higher squeeze pressures, and the pressure effect will easily win.**The diffusion coefficient?**For small air molecules, diffusion coefficients are surprisingly similar between typical adhesives, in the region of 10^{-8}m²/s. Presumably highly entangled, highly viscous, highly filled adhesives have somewhat lower values. Again, these systems will need higher squeeze pressures so things might balance out.**The solubility of the air in the adhesive?**This is governed by the Henry constant. Double the solubility and the time to collapse is halved. How do you know the Henry constant? There are two problems here - knowing the value and having the right units. The Gogos paper (below) uses the "solubility" form of Henry which is the inverse of the "volatility" form that is commonly quoted. So even if you have a value for H it might be the inverse of the one used here, and might be shown as K. Then we have a choice on the top of mole, g, kg and on the bottom m³, cm³ and of atm, Pa, kPa etc. Then there are the "dimensionless" Henry constants. If you find a value for Henry for air in your adhesive, the chances are very small that it will be the right way up and the right units. For example, Gogos quotes a value of 0.0966 kg/m³atm which cannot be plugged directly into his formula (below) because the dimensions are wrong. You have to convert it to mol/m³Pa because his equation features R in J/mol.K. That's why the input is in the units used by Kontopoulou. Values I have found are: Polyethylene: 2.2e-5, Rubber: 3.6e-5, Polystyrene: 6.8e-5, PDMS: 5.8e-5, PU: <1e-6, PMMA: <1e-6. The values for PU and PMMA are a warning that more polar adhesives have a much lower air solubility so getting rid of bubbles will be harder - so experiments in silicones might be deceptive!**The temperature?**Over any practical range for our adhesives, the effect is not massive. Because the formula involves T in °K, switching from 20°C to 100°C is a change of 293/393, only a 25% reduction in time. However, H does change with T so you have an extra complication.**The fact that the bubbles become cylinders once they are in contact with the adherend surfaces?**When bubbles are still bubbles, there is a standard correction factor of ~0.7 to take into account the fact that air can no longer diffuse equally in all directions. If the bubble nears two surfaces then presumably the overall factor is 0.5. So bubbles that are trapped might take 2x longer than the simple theory suggests. This has still avoided the direct question. A theory for a cylindrical bubble is beyond my capabilities.

### The formula

An excellent numerical scheme from M Kontopoulou and J Vlachopoulos, *Bubble Dissolution in Molten Polymers and Its Role in Rotational Molding*, Polym. Eng. Sci.,1999,39, 1189-1198, is a gold standard paper, validated by the key data (used by everyone) from Spence. However, as shown by George Gogos, *Bubble Removal in Rotational Molding*, Polym. Eng. Sci., 2004, 44, 388-394, although the Kontopoulou approach is excellent, a simple equation is good enough for most of us, especially given the uncertainties of the core parameters.

Gogos tells us that for a bubble of diameter D_{o} of air with diffusion coefficient D with a Henry constant H in the adhesive, with an excess pressure ΔP, and taking into account RT (gas constant times T assumed here to be 25°C) the time t_{Gone} for the bubble to disappear is given by:

`t_"Gone"=1/(8DHRT)(1+ΔP)^(1/3)/(ΔP)D_o^2`

Because of the squared dependence on D_{0} it's clear that smaller bubbles disappear much faster. D and H have linear effects and the ΔP effect starts off large, gradually losing its potency, so we don't have to squeeze *all* that hard!

Why the squared dependence on D_{0}? In Gogos it arises from his version of the pressure dependence of diameter. We might equally say that diffusion is through the bubble surface area which scales as the square of the diameter.

### It's OK in theory but ...

We all need to educate ourselves on the theory and its implications, so we need to validate a reasonable set of assumptions for our own system. So. Set up your system so you can put a nice blob of adhesive onto a glass surface, inject a bubble into the adhesive and, with your video camera *beneath* the glass, capture that moment because the needle is a convenient reference size for your image. Now gently place another piece of (tough) glass on top so you start with a fairly circular blob in your microscope. Now apply a known force, either from an Instron-like device or just a known mass m on top (that's why we look from underneath). At any moment you know the area of the adhesive and from m.g you know the force, hence you know the pressure at any moment.The video can follow the reduction in bubble size then, with a few reasonable approximations you should be able to link the data to the simulation.

There's a bonus. Many people don't believe that the bubble really dissolves into the bulk, so they expect it to bounce back if the pressure is removed. OK, check for yourself. At any moment, remove the weight from the top and see what happens to the bubble...

### But remember ...

If you *can* degas your adhesive this will speed up bubble removal. As a rule of thumb, a totally degassed system is equivalent to a ΔP of ~1 bar (which makes some sort of sense as there's now a 1 bar concentration gradient), a useful boost to speed.