**4.1 Nucleation of droplets**

We could think that the more straightforward way to form a cloud droplet would be by condensation in a saturated environment, when some water molecules collide by chance to form a cluster that will further grow to a droplet by picking up more and more molecules from the vapour phase. This process is termed *homogeneous nucleation*. The survival and further growth of the droplet in its environment will depend on whether the Gibbs free energy of the droplet and its surrounding will decrease upon further growth. We note that,

Atmospheric Thermodynamics 67

The greater *e* with respect to *es*, that is the degree of supersaturation, the smaller the radius

It can be shown from (48) that a droplet with a radius as small as 0.01 μm would require a supersaturation of 12% for getting activated. However, air is seldom more than a few percent supersaturated, and the homogeneous nucleation process is thus unable to explain the generation of clouds. Another process should be invoked: the *heterogeneous nucleation*. This process exploit the ubiquitous presence in the atmosphere of particles of various nature (Kaufman et al., 2002), some of which are soluble (hygroscopic) or wettable (hydrophilic) and are called *Cloud Condensation Nuclei (CCN)*. Water may form a thin film on wettable particles, and if their dimension is beyond the critical radius, they form the nucleus of a droplet that may grow in size. Soluble particles, like sodium chloride originating from sea spray, in presence of moisture absorbs water and dissolve into it, forming a droplet of solution. The saturation vapour pressure over a solution is smaller than over pure water,

� = ��

Where *e* in the vapour pressure over pure water, and *e'* is the vapour pressure over a solution containing a *mole fraction f* (number of water moles divided by the total number of

Let us consider a droplet of radius *r* that contains a mass *m* of a substance of molecular weight *Ms* dissolved into *i* ions per molecule, such that the effective number of moles in the solution is *im/Ms* . The number of water moles will be *((4/3)πr3ρ - m)/Mw* where *ρ* and *Mw* are

Eq. (49) and (50) allows us to express the reduced value *e'* of the saturation vapour pressure for a droplet of solution. Using this result into (48) we can compute the saturation vapour

> ����� �� � ���� ��� � � ������� � ��

The plot of supersaturation *e'/es* -1 for two different values of *m* is shown in fig. 8, and is

Figure 8 clearly shows how the amount of supersaturation needed to sustain a droplet of solution of radius *r* is much lower than what needed for a droplet of pure water, and it decreases with the increase of solute concentration. Consider an environment supersaturation of 0.2%. A droplet originated from condensation on a sphere of sodium chloride of diameter 0.1 μm can grow indefinitely along the blue curve, since the peak of the curve is below the environment supersaturation; such droplet is *activated*. A droplet originated from a smaller grain of sodium chloride of 0.05 μm diameter will grow until

= �� � ���� ��� � � ������� � ��

the water density and molecular weight respectively. The water mole fraction *f* is:

� (48)

� (49)

(50)

(51)

�� <sup>=</sup> �� ������� � ��

beyond which droplets become activated.

and the fractional reduction is given by *Raoult's law*:

� =

� � �������� ��

� � �������� �� ��� ��

pressure in equilibrium with a droplet of solution of radius *r*:

= ��� � ��

�� ��

moles) of pure water.

named *Köhler curve*.

by creating a droplet, work is done not only as expansion work, but also to form the interface between the droplet and its environment, associated with the *surface tension* at the surface of the droplet of area *A*. This originates from the cohesive forces among the liquid molecules. In the interior of the droplet, each molecule is equally pulled in every direction by neighbouring molecules, resulting in a null net force. The molecules at the surface do not have other molecules on all sides of them and therefore are only pulled inwards, as if a force acted on interface toward the interior of the droplet. This creates a sort of pressure directed inward, against which work must be exerted to allow further expansion. This effect forces liquid surfaces to contract to the minimal area.

Let *σ* be the energy required to form a droplet of unit surface; then, for the heterogeneous system droplet-surroundings we may write, for an infinitesimal change of the droplet:

$$dG = -SdT + Vdp + (\mu\_\nu - \mu\_l)dm\_\nu + \sigma dA \tag{46}$$

We note that *dmv = - dml = - nldV* where *nl* is the number density of molecules inside the droplet. Considering an isothermal-isobaric process, we came to the conclusion that the formation of a droplet of radius *r* results in a change of Gibbs free given by:

$$
\Delta G = 4\pi r^2 \sigma - \frac{4}{3} \pi r^3 n\_l KT \ln\left(\frac{e}{e\_5}\right) \tag{47}
$$

Where we have used (36). Clearly, droplet formation is thermodynamically unfavoured for *e < es*, as should be expected. If *e > es*, we are in supersaturated conditions, and the second term can counterbalance the first to give a negative *ΔG*.

Fig. 7. Variation of Gibbs free energy of a pure water droplet formed by homogeneous nucleation, in a subsaturated (upper curve) and a supersaturated (lower curve) environment, as a function of the droplet radius. The critical radius *r0* is shown.

Figure 7 shows two curves of *ΔG* as a function of the droplet radius *r*, for a subsaturated and supersaturated environment. It is clear that below saturation every increase of the droplet radius will lead to an increase of the free energy of the system, hence is thermodynamically unfavourable and droplets will tend to evaporate. In the supersaturated case, on the contrary, a critical value of the radius exists, such that droplets that grows by casual collision among molecules beyond that value, will continue to grow: they are said to get *activated*. The expression for such *critical radius* is given by the *Kelvin's formula*:

by creating a droplet, work is done not only as expansion work, but also to form the interface between the droplet and its environment, associated with the *surface tension* at the surface of the droplet of area *A*. This originates from the cohesive forces among the liquid molecules. In the interior of the droplet, each molecule is equally pulled in every direction by neighbouring molecules, resulting in a null net force. The molecules at the surface do not have other molecules on all sides of them and therefore are only pulled inwards, as if a force acted on interface toward the interior of the droplet. This creates a sort of pressure directed inward, against which work must be exerted to allow further expansion. This effect forces

Let *σ* be the energy required to form a droplet of unit surface; then, for the heterogeneous system droplet-surroundings we may write, for an infinitesimal change of the droplet:

We note that *dmv = - dml = - nldV* where *nl* is the number density of molecules inside the droplet. Considering an isothermal-isobaric process, we came to the conclusion that the

�

Where we have used (36). Clearly, droplet formation is thermodynamically unfavoured for *e < es*, as should be expected. If *e > es*, we are in supersaturated conditions, and the second

Fig. 7. Variation of Gibbs free energy of a pure water droplet formed by homogeneous nucleation, in a subsaturated (upper curve) and a supersaturated (lower curve) environment, as a function of the droplet radius. The critical radius *r0* is shown.

*activated*. The expression for such *critical radius* is given by the *Kelvin's formula*:

Figure 7 shows two curves of *ΔG* as a function of the droplet radius *r*, for a subsaturated and supersaturated environment. It is clear that below saturation every increase of the droplet radius will lead to an increase of the free energy of the system, hence is thermodynamically unfavourable and droplets will tend to evaporate. In the supersaturated case, on the contrary, a critical value of the radius exists, such that droplets that grows by casual collision among molecules beyond that value, will continue to grow: they are said to get

��������� � �

formation of a droplet of radius *r* results in a change of Gibbs free given by:

�� � ����� � �

�� � ���� � ��� � (�� � ��)��� � ��� (46)

���

� (47)

liquid surfaces to contract to the minimal area.

term can counterbalance the first to give a negative *ΔG*.

$$
\sigma\_0 = \frac{2\sigma}{n\_l kT \ln\left(\frac{\sigma}{a\_s}\right)}\tag{48}
$$

The greater *e* with respect to *es*, that is the degree of supersaturation, the smaller the radius beyond which droplets become activated.

It can be shown from (48) that a droplet with a radius as small as 0.01 μm would require a supersaturation of 12% for getting activated. However, air is seldom more than a few percent supersaturated, and the homogeneous nucleation process is thus unable to explain the generation of clouds. Another process should be invoked: the *heterogeneous nucleation*. This process exploit the ubiquitous presence in the atmosphere of particles of various nature (Kaufman et al., 2002), some of which are soluble (hygroscopic) or wettable (hydrophilic) and are called *Cloud Condensation Nuclei (CCN)*. Water may form a thin film on wettable particles, and if their dimension is beyond the critical radius, they form the nucleus of a droplet that may grow in size. Soluble particles, like sodium chloride originating from sea spray, in presence of moisture absorbs water and dissolve into it, forming a droplet of solution. The saturation vapour pressure over a solution is smaller than over pure water, and the fractional reduction is given by *Raoult's law*:

$$f = \frac{e'}{e} \tag{49}$$

Where *e* in the vapour pressure over pure water, and *e'* is the vapour pressure over a solution containing a *mole fraction f* (number of water moles divided by the total number of moles) of pure water.

Let us consider a droplet of radius *r* that contains a mass *m* of a substance of molecular weight *Ms* dissolved into *i* ions per molecule, such that the effective number of moles in the solution is *im/Ms* . The number of water moles will be *((4/3)πr3ρ - m)/Mw* where *ρ* and *Mw* are the water density and molecular weight respectively. The water mole fraction *f* is:

$$f = \frac{\frac{\left(\frac{4}{3}\pi r^3 \rho - m\right)}{M\_W}}{\frac{\left(\frac{4}{3}\pi r^3 \rho - m\right)}{M\_W} - \frac{tm}{M\_S}} = \left(1 + \frac{lmM\_w}{M\_s \left(\frac{4}{3}\pi r^3 \rho - m\right)}\right)^{-1} \tag{50}$$

Eq. (49) and (50) allows us to express the reduced value *e'* of the saturation vapour pressure for a droplet of solution. Using this result into (48) we can compute the saturation vapour pressure in equilibrium with a droplet of solution of radius *r*:

$$\frac{e'}{e\_s} = \exp\left(\frac{2\sigma}{nKTr}\right) \left(1 + \frac{im\mathcal{M}\_w}{\mathcal{M}\_s \left(\frac{4}{3}\pi r^3 \rho - m\right)}\right)^{-1} \tag{51}$$

The plot of supersaturation *e'/es* -1 for two different values of *m* is shown in fig. 8, and is named *Köhler curve*.

Figure 8 clearly shows how the amount of supersaturation needed to sustain a droplet of solution of radius *r* is much lower than what needed for a droplet of pure water, and it decreases with the increase of solute concentration. Consider an environment supersaturation of 0.2%. A droplet originated from condensation on a sphere of sodium chloride of diameter 0.1 μm can grow indefinitely along the blue curve, since the peak of the curve is below the environment supersaturation; such droplet is *activated*. A droplet originated from a smaller grain of sodium chloride of 0.05 μm diameter will grow until

Atmospheric Thermodynamics 69

Where we have used the ideal gas equation for water vapour. We should think of *e(r)* as given by *e'* in (49), but in fact we can approximate it with the saturation vapour pressure over a plane surface *es*, and pose *(e(∞)-e(r))/e(∞)* roughly equal to the supersaturation

> � = ���(�) ��

This equation shows that the radius growth is inversely proportional to the radius itself, so that the rate of growth will tend to slow down with time. In fact, condensation alone is too slow to eventually produce rain droplets, and a different process should be invoked to

The droplet of density *ρl* and volume *V* is suspended in air of density *ρ* so that under the effect of the gravitational field, three forces are acting on it: the gravity exerting a downward force *ρ<sup>l</sup> V*g , the upward Archimede's buoyancy *ρV* and the *drag force* that for a sphere, assumes the form of the *Stokes' drag 6πηrv* where *η* is the viscosity of the air and *v* is the steady state terminal fall speed of the droplet. In steady state, by equating those forces and assuming the droplet density much greater than the air, we get an expression for the

> � � �����

Such speed increases with the droplet dimension, so that bigger droplets will eventually collide with the smaller ones, and may entrench them with a *collection efficiency E* depending on their radius and other environmental parameters , as for instance the presence of electric fields. The rate of increase of the radius *r1* of a spherical collector drop due to collision with water droplets in a cloud of *liquid water content wl* , that is is the mass density

Since *v1* increases with *r1*, the process tends to speed up until the collector drops became a rain drop and eventually pass through the cloud base, or split up to reinitiate the

A cloud above 0° is said a *warm cloud* and is entirely composed of water droplets. Water droplet can still exists in *cold clouds* below 0°, although in an unstable state, and are termed *supecooled*. If a cold cloud contains both water droplets and ice, is said *mixed cloud*; if it

For a droplet to freeze, a number of water molecules inside it should come together and form an *ice embryo* that, if exceeds a critical size, would produce a decrease of the Gibbs free energy of the system upon further growing, much alike the homogeneous condensation from the vapour phase to form a droplet. This glaciations process is termed *homogeneous freezing*, and below roughly -37 °C is virtually certain to occur. Above that temperature, the

<sup>≅</sup> ����� ���

�� <sup>=</sup> (�����)��� ���

� (55)

� (56)

≅ (57)

��

create droplet with radius greater than few tens of micrometers.

*S=(e(∞)-es)/es* to came to:

**4.3 Collision and coalescence** 

terminal fall speed:

process.

��

� =

of liquid water in the cloud, is given by:

��

**4.4 Nucleation of ice particles** 

contains only ice, it is said *glaciated*.

when the supersaturation adjacent to it is equal to the environmental: attained that maximum radius, the droplet stops its grow and is in stable equilibrium with the environment. Such *haze dropled* is said to be *unactivated*.

Fig. 8. Kohler curves showing how the critical diameter and supersaturation are dependent upon the amount of solute. It is assumed here that the solute is a perfect sphere of sodium chloride (source: http://en.wikipedia.org/wiki/Köhler\_theory).

#### **4.2 Condensation**

The droplet that is able to pass over the peak of the Köhler curve will continue to grow by condensation. Let us consider a droplet of radius *r* at time *t*, in a supersaturated environment whose water vapour density far from the droplet is *ρv(∞)*, while the vapour density in proximity of the droplet is *ρv(r)* . The droplet mass *M* will grow at the rate of mass flux across a sphere of arbitrary radius centred on the droplet. Let *D* be the diffusion coefficient, that is the amount of water vapour diffusing across a unit area through a unit concentration gradient in unit time, and *ρv(x)* the water vapour density at a distance *x > r* from the droplet. We will have:

$$\frac{d\mathbf{M}}{dt} = 4\pi\mathbf{x}^2 D \frac{d\rho\_v(\mathbf{x})}{dx} \tag{52}$$

Since in steady conditions of mass flow this equation is independent of *x*, we can integrate it for *x* between *r* and *∞* to get:

$$\frac{d\mathcal{M}}{dt}\int\_{r}^{\infty}\frac{d\boldsymbol{x}}{\boldsymbol{x}^{2}} = \int\_{\rho\_{\boldsymbol{\nu}}(r)}^{\rho\_{\boldsymbol{\nu}}(\infty)} d\rho\_{\boldsymbol{\nu}}(\boldsymbol{x})\tag{53}$$

Or, expliciting *M* as *(4/3)πr3ρl*:

$$\frac{dr}{dt} = \frac{D}{r\rho\_l} \left(\rho\_\nu(\infty) - \rho\_\nu(r)\right) = \frac{D\rho\_\nu(\infty)}{r\rho\_l e(\infty)} \left(e(\infty) - e(r)\right) \tag{54}$$

when the supersaturation adjacent to it is equal to the environmental: attained that maximum radius, the droplet stops its grow and is in stable equilibrium with the

Fig. 8. Kohler curves showing how the critical diameter and supersaturation are dependent upon the amount of solute. It is assumed here that the solute is a perfect sphere of sodium

The droplet that is able to pass over the peak of the Köhler curve will continue to grow by condensation. Let us consider a droplet of radius *r* at time *t*, in a supersaturated environment whose water vapour density far from the droplet is *ρv(∞)*, while the vapour density in proximity of the droplet is *ρv(r)* . The droplet mass *M* will grow at the rate of mass flux across a sphere of arbitrary radius centred on the droplet. Let *D* be the diffusion coefficient, that is the amount of water vapour diffusing across a unit area through a unit concentration gradient in unit time, and *ρv(x)* the water vapour density at a distance *x > r*

= ����� ���(�)

� <sup>=</sup> � ���(�) ��(�)

���(∞) � ��(�)� = ���(�)

Since in steady conditions of mass flow this equation is independent of *x*, we can integrate it

�� (52)

��(�) (53)

����(�) ��(∞) � �(�)� (54)

chloride (source: http://en.wikipedia.org/wiki/Köhler\_theory).

�� ��

�� �� � �� �� �

**4.2 Condensation** 

from the droplet. We will have:

for *x* between *r* and *∞* to get:

Or, expliciting *M* as *(4/3)πr3ρl*:

�� �� <sup>=</sup> � ���

environment. Such *haze dropled* is said to be *unactivated*.

Where we have used the ideal gas equation for water vapour. We should think of *e(r)* as given by *e'* in (49), but in fact we can approximate it with the saturation vapour pressure over a plane surface *es*, and pose *(e(∞)-e(r))/e(∞)* roughly equal to the supersaturation *S=(e(∞)-es)/es* to came to:

$$\frac{dr}{dt}r = \frac{\mathcal{D}\rho\_V(\alpha)}{\rho\_I}\mathcal{S}\tag{55}$$

This equation shows that the radius growth is inversely proportional to the radius itself, so that the rate of growth will tend to slow down with time. In fact, condensation alone is too slow to eventually produce rain droplets, and a different process should be invoked to create droplet with radius greater than few tens of micrometers.

### **4.3 Collision and coalescence**

The droplet of density *ρl* and volume *V* is suspended in air of density *ρ* so that under the effect of the gravitational field, three forces are acting on it: the gravity exerting a downward force *ρ<sup>l</sup> V*g , the upward Archimede's buoyancy *ρV* and the *drag force* that for a sphere, assumes the form of the *Stokes' drag 6πηrv* where *η* is the viscosity of the air and *v* is the steady state terminal fall speed of the droplet. In steady state, by equating those forces and assuming the droplet density much greater than the air, we get an expression for the terminal fall speed:

$$\upsilon = \frac{2}{9} \frac{\rho\_l g r^2}{\eta} \tag{56}$$

Such speed increases with the droplet dimension, so that bigger droplets will eventually collide with the smaller ones, and may entrench them with a *collection efficiency E* depending on their radius and other environmental parameters , as for instance the presence of electric fields. The rate of increase of the radius *r1* of a spherical collector drop due to collision with water droplets in a cloud of *liquid water content wl* , that is is the mass density of liquid water in the cloud, is given by:

$$\frac{dr}{dt} = \frac{(v\_1 - v\_2)\nu\_l E}{4\rho\_l} \cong \frac{v\_1 \nu\_l E}{4\rho\_l} \cong \tag{57}$$

Since *v1* increases with *r1*, the process tends to speed up until the collector drops became a rain drop and eventually pass through the cloud base, or split up to reinitiate the process.

#### **4.4 Nucleation of ice particles**

A cloud above 0° is said a *warm cloud* and is entirely composed of water droplets. Water droplet can still exists in *cold clouds* below 0°, although in an unstable state, and are termed *supecooled*. If a cold cloud contains both water droplets and ice, is said *mixed cloud*; if it contains only ice, it is said *glaciated*.

For a droplet to freeze, a number of water molecules inside it should come together and form an *ice embryo* that, if exceeds a critical size, would produce a decrease of the Gibbs free energy of the system upon further growing, much alike the homogeneous condensation from the vapour phase to form a droplet. This glaciations process is termed *homogeneous freezing*, and below roughly -37 °C is virtually certain to occur. Above that temperature, the

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critical dimensions of the ice embryo are several micrometers, and such process is not favoured. However, the droplet can contain impurities, and some of them may promote collection of water droplets into an ice-like structure to form a ice-like embryo with dimension already beyond the critical size for glaciations. Such particles are termed *ice nuclei* and the process they start is termed *heterogeneous freezing*. Such process can start not only within the droplet, but also upon contact of the ice nucleus with the surface of the droplet (*contact nucleation*) or directly by deposition of ice on it from the water vapour phase (*deposition nucleation*). Good candidates to act as ice nuclei are those particle with molecular structure close to the hexagonal ice crystallography. Some soil particles, some organics and even some bacteria are effective nucleators, but only one out of 103-105 atmospheric particles can act as an ice nucleus. Nevertheless ice particles are present in clouds in concentrations which are orders of magnitude greater than the presence of ice nuclei. Hence, ice multiplication processes must be at play, like breaking of ice particles upon collision, to create ice splinterings that enhance the number of ice particles.
