Nucleation Continued: Requirements for
Nucleation Requirements
This blog is a continuation of two previous blogs:
- 1) Bubble Nucleation: Bubble stability: Jan 16 2017
- 2) Energetics of Nucleation: Jan 17 2017
I will repeat that much of this blog is taken from parts of a book I wrote (never published) a decade or so ago. Accordingly image numbers and equation numbers with adhere to the book in its current format.
In the bubble stability blog I discussed that in order for a bubble to be stable it had to be large enough that the liquid’s cohesive forces are not felt across the bubble, and more than likely by its third neighbor along the tensile layer. In most systems a microbubble is the smallest stable bubble.
Due to the above we realize that in order for a bubble to nucleate there must be a sufficient number of molecules in some sort of thermal contact that can change states from:
- a) Liquid state to vapor state : A vapor bubble
- b) Dissolved gas state into a gaseous state: A gaseous bubble
Similarly in order for a stable droplet to form it too must be a given size. The reasoning is due to the surface area defined tensile layer requiring positive work (endothermic) and the volume defined interior bulk of the droplet being being exothermic. One can envision a droplet growing so long as it is in thermal contact with an entity that it can extract energy from. Even so for the initial tensile layer formation there must be numerous molecules in some sort of thermal contact that can change states from vapor/gas state into liquid state.
In context of this blog, a cluster is an ensemble of molecules, which are both similar (not necessarily identical) and are quasi-bound. Quasi-bound implies that these molecules, can behave unison, thus allowing a particular nucleation event to occur. Quasi-bound implies from a thermodynamic perspective that they possess similar behavior at a given system’s temperature, and/or pressure.
Throughout this blog we shall consider:
- 1) A-cluster as an ensemble of aqueous molecules within a liquid
- 2) L-cluster as an ensemble of activated liquid molecules within a liquid
- 3) G-cluster as an ensemble of gaseous/vaporous molecules
It should be emphasized that the molecules of a cluster do not have to be identical. For example twoifferent types of aqueous molecules, will both want to attain volume (i.e. form a bubble) if they can, therefore they can be part of the same a-cluster.
According unlike evaporation or condensation, nucleation generally requires some form of cluster of molecules. We can further say: Nucleation involves the formation of a curved tensile layer therefore both a pressure change and internal energy change must occur. Furthermore, most nucleation processes involve a phase change hence, a change to the mean molecular volume of the nucleating cluster occurs. Ultimately, nucleation is a rare thermodynamic process wherein, volume, pressure and internal energy, are all variables.
As previously stated in blogs due the basic construct of traditional thermodynamics, it cannot deal with nucleation processes properly. Especially processes wherein both pressure and volume increases i.e. bubble nucleation.
Nucleation is the instantaneous, significant and uniform change of all the molecules within a cluster, into a state wherein all the cluster’s molecules now coexist exist at a different energy level, which is often a state of higher energy. Higher energy state means energy is required for nucleation. Most often the energy required for the cluster to nucleate, is extracted from its surrounding (neighboring) molecules.
Probability
I will state at this point that the extraction of energy from neighboring molecules is nothing new to thermodynamics. It is generally based upon probabilities and is traditionally dealt with using the Boltzmanns exponential. Sadly Boltzmann’s probabilities do not apply to nucleation processes. I may discuss the reasons that this is the case, but that would have to be another blogs (and I will have to learn how to write exponential equations in wordpress before I would ever attempt that.
I will also elaborate that Boltzmann’s exponential is wrongly applied to vaporization process. The basic reason is that you cannot write a a proper probability within clearly defining its path of energy transfer. I discuss this in my paper titled “Latent heat and critical temperature: A unique perspective” Physics Essays 2013. The paper can also be seen on my website: http://www.newthermodynamics.com and click on my papers.
The thing about nucleation is the paths are not as simple as they would be for simple evaporation. At this current time I believe that the best way to view this cascading probabilities something I discuss in my unpublished book. Sorry its not in the public view just yet.
Furthermore: You cannot write the probability for nucleation based upon the energy requirement of the whole bubble. Rather the probability is based upon energy required per molecule. This will be discussed in more detail later in this blog
I also want to say that the above concerning evaporation also means that we have to rewrite the Clausius-Clapeyron equation. This is actually done from first principles in my self-published book Changing our perspective Part 1: A new thermodynamics. For those of you who can wait (lol) I plan on republishing this book with corrections sometime this year when/if I find time. Sorry I have gotten of topic. Let us just leave it for now that nucleation processes generally initially require energy, hence are fundamentally enothermic processes, with a probability of occurrence that is based upon energy requirements. If you prefer there is a probability of the bubble or droplet accessing the energy required for nucleation, i.e. I sometimes call this the accessible energy (generally kT per neighboring molecule: Boltzmann at least got this right: or should I say I still believe this at this time).
We can further elaborate that there must be some sort of second probability that is not energy related but related to the probability of a cluster existing for a given time span that is sufficiently large for either a bubble or a droplet to nucleate.
Aqueous (dissolved gas) to Gaseous Bubble
Consider bubbles that nucleate from an a-cluster. Assume that the aqueous molecules are not bound to the surrounding liquid, and that the bonding potential in the gaseous state is small then we can approximate the work required for a nucleating bubble as:
Wb=Ab@+PbVb 4.1.4
Where @ is the surface tension
Eqn 4.1.4 is for a bubble filled with an ideal gas. The work required to nucleate such a bubble, can be viewed as the summation of two positive terms:
- 1) Energy to form the tensile layer (positive): Ab@
- 2) Work required to create a volume:: PbVb
Doing a bit of mathematical analysis (in my unpublished book) and then collecting the terms, the work required to nucleate a bubble from an a-cluster becomes:
Wb=(4/3)Ab@ +VbPl 4.1.9
Note: The above analysis is only an approximation, based upon the molecules of the a-cluster having no bonding energy in both the liquid and gaseous state, i.e. an ideal gas.
Bubbles from L-clusters
Calculating the work required to nucleate a bubble from a l-cluster could be trickier than it was for an a-cluster, because there are binding potentials in both the liquid and vaporous states. We can write the energy required in the following form:
Wb=Ab@+Xl +Xcy(Tb-Tw) 4 .1.14
X is number of molecules, @ is surface tension, l is latent heat per molecule, cy is heat capacity per molecule
Since all the molecules must have the energy required to vaporize at the same instant, it is very likely that vapor bubbles only nucleate when the X molecules of the l-cluster are at, or above their boiling point. Furthermore, a liquid’s boiling point depends upon the system’s pressure, and it is at the pressure within the bubble that we are concerned. Therefore, the X molecule’s boiling point must depend upon the resulting bubble’s internal pressure, which depends upon the resulting bubble’s radius, which depends upon the value of X, as well as our system’s temperature. We arrive at the bizarre realization that two different sizes of nucleating bubbles, in a given system will possess different boiling points (Tb at Pb).
Although the energy required for nucleation is an increasing function of radius, it must be stated that the energy required per molecule is a decreasing function of radius. Remember probability is based upon energy required per molecule. In other words a larger bubble has a higher probability of nucleation from an energy perspective. Of course the probability of a cluster existing decreases as the cluster size increases. I am sorry but this all belongs in a book and not blog, but herein I am just trying to provide you with some of the basic understanding:
- 1) A bubble has to be large enough to be stable. Hence can only form from a large enough cluster. Of course the lower the pressure then the smaller the cluster size can be in order to nucleate a stable bubble of a given size
- 2) Larger cluster size means the lower the energy required per molecule and the higher the basic probability (based upon energy per molecule) will be but perhaps the lower the cascading probability actually is (sorry but it is not all that simple: I am trying to make it as simple as possible)
- 3) Aqueous clusters require less energy to nucleate because there is little or no bonding between dissolved gas and the surrounding liquid
- 4) Vaporous bubbles like bubbles in water require the same work as aqueous clusters plus the energy required to break bonds and elevate the temperatures within: superheat if you prefer.
If you consider all the possible variables in my short list above and then superimpose the blatant faults of traditional thermodynamics, one actually begins to understand why we made a mess of the science of nucleation for so. The ramifications of what I say starts with understanding boiling process, cavitations and for me ends with actually understanding decompression illness (DCI) in scuba diving [okay this is where I actually started: Of course no hyperbaric doctor would listen to me : call it human arrogance, which may lend itself to be called human nature (I am being nice)].
Crevices and Surface Bubbles
We certainly expect that bubbles, which nucleate in the middle of some liquid, will be spherical and the nucleation of such bubbles is commonly referred to as homogeneous bubble nucleation. However, non-spherical bubbles do nucleate in crevices, and upon surfaces in what is commonly referred to as heterogeneous bubble nucleation.
Crevices
Assume that the crevice is a perfect cone of radius R, as is shown in Fig. 4.6.1. Consider the simplest type of bubble that being an a-cluster nucleating into a gas filled bubble. The work required to nucleate for such a conical section of a bubble is defined by eqn 4.1.4:
It can be readily shown (in my unpublished book) ,that if the tensile layers of both the cone and the bubble have identical curvature then the pressure change across these tensile layers must be the same for both. Therefore, the work required per molecule is equal for both a cone and a bubble of the same radius. If the work required per molecule (Wb/X) is equal, then crevices simply allow for smaller clusters (a-cluster, or l-cluster) to nucleate into bubbles, than would otherwise be possible.
Surface Bubbles
Similar concepts can be applied to a flat or semi-flat surface. Bubbles often nucleate as half spheres upon flat surfaces. It can be easily shown (in my book): An a-cluster only has to be ½ the size to heterogeneously nucleate as a half sphere, as it has to be to nucleate as a same radii full sphere. Since the work required per molecule, hence probability of nucleation, is the same for a ½ bubble as a full bubble of the same radius, just like crevices, flat surfaces allow for clusters (a-cluster or l-cluster) to nucleate heterogeneously when the cluster is too small to nucleate homogeneously.
Certainly, we must also consider that larger radii ½ bubbles would have a more stable tensile layer than a smaller radii full bubble, which has the same number of gaseous molecules inside.
Conclusions
Previous theories for bubbles preferentially nucleating in crevices are based upon crevices requiring a smaller tensile layer than a spherical bubble thus less total work and the contact angles between the interface and crevice surface. Such theories are hard to visualize and wrong. Correlations between contact angle should be treated as a secondary phenomenon or a result.
When all is said and done, our explanation as to why bubbles preferentially nucleate in crevices is easier to rationalize than previous theories ever were.
Short Discussion of Droplets
Ignoring the work associated with the pressure increase inside of the droplet, then the work required to nucleate a droplet, would be:
Wd=Ad@-Xl
It can be further shown the following
Work Required to Nucleate a Droplet
The three factors that must be added together to calculate the energy required to nucleate a liquid droplet are:
- 1) Energy to form tensile layer (positive):
- 2) Work to increase the droplet’s internal pressure (positive):
- 3) Latent heat of condensation (negative):
Accordingly (after a bunch of math in my unpublished book), the work required to nucleate a droplet becomes:
Wd=2Ad@-Ul 6.1.10
Ul is the total bonding energy in the liquid state which is a function of the number of molecules (X), and binding energy per molecule (ul). Therefore, we can rewrite 6.1.10, as:
Wd=2Ad@-Xul 6.1.11
Anyhow after a bunch of analysis (again in my unpublished book) we arrive at:
It is interesting that we can simply calculate the radius of a large zero work droplet, by multiplying the critical radius by: 3/2. The true importance of the critical radius is this: For droplets that are smaller than the critical radius, it will take an increasing amount of energy per unit radius for the droplet to grow. For droplets, that are larger than the critical radius, it will take a decreasing amount of energy per unit radius, for the droplet to grow. From a purely energetics perspective a critical radius droplet, is equally likely to grow or decay.
Understandably, as the droplet’s radius (rd) increases, then the surface area to volume ratio decreases, and the work required per molecule decreases. Hence the basic probability of nucleation increases (probability based upon work required per molecule) but the larger the g-cluster then the lower the probability (likelihood) of all the molecules being in thermal contact.
The reality becomes: Only extremely small g-clusters nucleating can exist hence only extremely small droplets nucleate. Moreover, such extremely small radii droplets must require energy to nucleate. Obviously, droplets must grow slowly from extremely small droplets into larger ones. Certainly, as the droplet grows, a decreasing amount of energy per molecule is going to be required, until the droplet eventually reaches the size of a zero energy droplet, as defined by eqn 1.6.23. After that point then the act of gaseous molecules condensing, become similar to condensation through a flat tensile layer, i.e. an exothermic process.
So where does the work required for droplet nucleation come from? We begin by considering that a g-cluster nucleates into a droplet, upon a surface. Now, the energy required for nucleation, has a probability of being accessed from the flat surface, as is shown in Fig. 6.2.4. i.e. the flat surface is acting as a heat bath (sink). The extracted energy (Q) from the flat solid surface must result in a temperature drop of that flat surface.
The same principle applies to raindrops. It is accepted that rain drops need something, i.e. a dust particle, to nucleate around. In Fig.6.2.5, is illustrated a g-cluster interacting with a dust particle on the L.H.S.. In order for a raindrop to nucleate, the dust particle must act as a heat sink, in much the same way as the flat surface did. Thus the nucleus, or nuclei, of a raindrop is nothing more than something, which can act as a heat sink. Plain and simple!
So we can now understand why clouds are formed around microscopic particles, such as salt crystals, dust and smoke. Experience has shown us, that without such nuclei, there would be no rain clouds. It is now understandable why: Cloud seeding generally uses hydroscopic particles (i.e. salts) sprayed out of airplanes to aid in cloud formation.
On a flat solid surface, the droplet will nucleate as a half droplet (half sphere). The work required per molecule, will be the same, as an equal radius full sized droplet, so long as the liquid has an affinity to the hard surface. Naturally, half as many molecules would be needed to attain the same critical radius and zero work radii, as a full sphere droplet.
If the flat solid surface is horizontal, and droplets continue to nucleate upon this surface, then eventually the half droplets will coalesce. As they do, the liquid’s surface area to volume ratio will decrease. Eventually, if enough coalesce then a thin film forms Now the energy by each gaseous molecule becoming part of the thin film of liquid is given by the latent heat of condensation.
It is accepted, that the more thermally conductive the surface is, the higher the droplet nucleation rate will be. Merte 1 (pg 233) states “It is noted that a 20 fold increase in thermal conductivity” of the flat surface, then “ the condensing coefficient increases fivefold”. This certainly jives well with our requirement of a heat sink for droplet nucleation. The higher the surface’s thermal conductivity then the better the heat sink it will be. Thus the best heat sinks must be materials with high thermal conductivity, such as metallic surfaces.
1H. Merte, Advances in Heat Transfer, l9, 181 (1973)