## Khler theory

Combining the Kelvin and Raoult effects, we get an expression for the ratio of the saturated vapour pressure over a flat surface of pure water, as derived in Section 5.1, to the saturated vapour pressure of a solution in a spherical droplet of radius r at temperature T. This saturation ratio is:

with, for water drops in air, a = ^ 3.3 x 10-7Km, (7.29a)

Rv b = ZvißiiJ^ ^ 4.3 x 10-6 (tMs/ßs) m3. (7.29b)

A plot of S as a function of radius is called a Köhler curve. Figure 7.3 shows an example of a Köhler curve highlighting the Kelvin and Raoult effects. The Köhler curve defines at what relative humidity the droplet would be in equilibrium with the environment.

The maximum of the Köhler curve plays an important role for droplet growth. The radius at the maximum is called the activation radius, denoted r*, and the value of the supersaturation at the activation radius is called the activation saturation ratio or critical saturation ratio, denoted S*.

The values of r* and S* follow from setting dS/dr = 0 in the Köhler equation. An exact calculation requires finding the root of a cubic polynomial. However, from the above graph it can be seen that both the Kelvin and Raoult effects are typically quite small around the activation radius (of order one percent). This suggests that we can use a Taylor expansion for both effects and find an approximate result. To do this, define a non-dimensional radius x as

V a and a non-dimensional parameter e as

We can write the Köhler curve equation, Eq. 7.28, as

For typical atmospheric values of the parameters a, b, and T it can be verified that e is quite small. Now assuming that x is not much bigger or much smaller than 1 - to be verified in hindsight - we can use a Taylor expansion for S in small e, e e i

x x3

We thus have dS e e

dx x2 x4

Setting this to zero gives the non-dimensional activation radius x*, x* + O(e), (7.35)

which is indeed not much larger than 1, validating the Taylor expansion.38 The corresponding critical saturation ratio is

38A full solution of dS/dx = 0 without approximation leads to

The difference between this exact expression for x* and the approximated expression is small; it can be shown that x* = V3 — e/6 + O(e2).

2 100

2 100

Figure 7.4 Köhler curves for different amounts of NaCl solute (in 10—16 g). In the limit of vanishing solute amount, the Köhler curve approaches that of the Kelvin effect (thin line).

Figure 7.4 Köhler curves for different amounts of NaCl solute (in 10—16 g). In the limit of vanishing solute amount, the Köhler curve approaches that of the Kelvin effect (thin line).

So the non-dimensional parameter e is a measure of the critical supersaturation.

Substituting the definitions of x and e, we find the activation radius r*

The critical supersaturation, defined as S* — 1, decreases when b increases, which is to say when the amount of solute increases in the droplet. At the same time the activation radius r* increases. Figure 7.4 shows various Köhler curves for different amounts of solute, illustrating the dependencies of r* and S* on the solute amount. Note that the activation radius is always very small. For any drops larger than about 1 |xm both the Kelvin and Raoult effects can be ignored.

The Koöhler curve contains a lot of information about initial cloud droplet growth. Consider Figure 7.5, where the Köhler curve and the level of the critical saturation ratio divides the figure into three different areas, labelled A, H, and N.

and the critical saturation ratio S1

Figure 7.5 Three distinct regions, A, H, and N, in the (r, RH) space, as defined by the Köhler curve.

Figure 7.5 Three distinct regions, A, H, and N, in the (r, RH) space, as defined by the Köhler curve.

Now suppose the nucleus has attracted some water and forms a small droplet of a particular radius r. Further suppose this droplet is moved into air of a particular relative humidity RH. This initial droplet then finds itself at the point (r, RH) in the figure and will be in one of the labelled areas.

Suppose the initial droplet was somewhere in area H. In this area the droplet is in an environment with a higher saturation ratio than the equilibrium saturation ratio for the droplet, as given by the Köhler curve. This means that the droplet is in a supersaturated environment and water vapour will start to condense onto the drop. The drop will grow; the point (r, RH) in the figure will move to the right, as indicated by the arrow. This growth will continue until the point hits the Köhler curve, where RH = S; here the droplet is in equilibrium with its environment and will remain steady. Such droplets are called haze.

Suppose the initial droplet was somewhere in area N. In this area the droplet is in an environment with lower relative humidity than the equilibrium saturation ratio for the droplet. The droplet will therefore start to evaporate and reduce in radius. The point (r, RH) for the droplet will move to the left in the figure until it hits the Koöhler curve, where the drop is again in equilibrium, RH = S. The droplet has again become haze. To find initial droplets in area N requires the existence of relatively large drops in subsat-urated air, which is possible when turbulence is present or when raindrops fall in subsaturated air.

Haze droplets on the boundary between areas H and N, given by the Köhler curve below the activation radius, are stable: if for some reason a haze droplet were to move away from the boundary the tendency would be for the droplet to be pushed back to the boundary. Haze is a stable state for droplets. For larger condensation nuclei the steady radius of haze droplets can be up to several hundreds of nanometres, and therefore become visible.

Suppose the initial droplet was somewhere in area A. When considering cloud droplet growth from initial condensation nuclei we start at very small radii. So typically, droplets in area A have to be in an environment with a relative humidity higher than the critical saturation ratio S*. The droplet is in a supersaturated environment and will start to grow. But now the growing droplet cannot hit the Köhler curve at any point; it will continue to grow indefinitely and will eventually become a cloud drop. Droplets in region A are said to be activated.

For droplets to become activated and grow into cloud drops the environmental supersaturation has to be larger than the critical supersaturation. Typical critical supersaturations are less than a couple of percent. This is the reason why in the atmosphere the relative humidity will never exceed 100% by very much. With any temporary increase in relative humidity, haze droplets would activate and form clouds. These growing cloud droplets would reduce the vapour content of the air until a relative humidity of 100% was achieved and droplets could no longer use up water vapour.

So at relative humidities below 100% any condensation nuclei will form stable haze. Suppose that this air is forced to lift by some external process; this will expand and cool down the air, thus increasing the relative humidity. Because of the stability of the haze part of the Köhler curve, the haze droplets move up on the Köhler curve and get a larger radius.39 This process

radius (|im) Figure 7.6 Growth of a haze drop in an updraft.

39The Kohler curve itself is a weak function of temperature. However, this does not change the main argument here.

continues until the air reaches critical supersaturation; the haze droplets will then have grown to the activation radius. Any further adiabatic cooling will activate the droplet and make it grow into cloud drops, and thus remove water vapour so as to keep the relative humidity down. Figure 7.6 shows this process on the Koöhler curve.

+1 0