So, now we can put the pieces together. The equilibrium vapor pressure for a water drop containing a solute is simply the triple product of the saturation vapor pressure for pure water over a plane surface, the curvature effect, and the solute effect:
This equation is called the Koehler Equation. It gives us the equilibrium vapor pressure that a drop will have for a given drop radius rd and given number of moles of solute Ns. We can see that there are two competing effects: the curvature or Kelvin effect that depends on the inverse of the drop radius, and the solute or Raoult effect that depends on the inverse of the drop radius cubed.
Remember we talked about the saturation ratio, S = e/es. We also talked about supersaturation, s = S – 1= e/es – 1. We define a new saturation ratio and supersaturation for a particle, Sk and sk, where “k” stands for “Koehler.”
sk is defined as follows:
Equation 5.13 is the form of the Koehler Equation that is most often used. Remember, this equation applies to individual drops. Each drop has its own Koehler curve because each drop has its own amount of solute of a given chemical composition. No two Koehler curves are alike, just as no two snowflakes are alike.
Let's look at a graph of this equation. At 20 oC, aK = 1.1 x 10–9 m, B = 4.3 x 10–6 m3 mole–1, and Ns is typically in the range of 10–18 to 10–15 moles.
Interpretation of the figure:
The video below (1:35) explains the Koehler Curve Equation in more depth:
Koehler theory is at the heart of cloud microphysics. It deals with two competing processes. One raises the equilibrium saturation vapor pressure above the [INAUDIBLE] saturation vapor pressure of the flat surface of your water. This is called the Kelvin effect, or the curvature effect. And the second process lowers the equilibrium vapor pressure. This is called Raoult's Law or the solute effect. Review the previous two sections of this lesson if you've forgotten these two effects. We can approximate the curvature effect as a constant over the drop radius and we can approximate the solute effect as the negative a constant over the drop radius cubed. Together they give us supersaturation for a drop. How do these two give us the Koehler curve? The curvature effect goes as a positive inverse of the radius. But the solute effect goes as the inverse of the radius cubed is negative. And at small r it is greater negative than the curvature effect is positive. As a drop gets bigger, then the curvature effect becomes more important. And then the drop equilibrium supersaturation follows the curvature effect. Note that each drop has its own Koehler curve. Supersaturation of the environment, which can be positive by radiative heating, cooling, mixing, [INAUDIBLE] descent, determines what will happen to the drop.
Note that the supersaturation is less than 0.2% for the smaller particle and less than 0.1% for the larger particle. As cooling occurs, which one will activate first?
ANSWER: The larger particle, because it has a lower critical supersaturation.
To see what happens to the drop in the atmosphere, we need to compare the Koehler curve for each drop to the ambient supersaturation of the environment. The Koehler curve is the equilibrium supersaturation, sk, for each drop and it varies as a function of the drop size. The ambient supersaturation, s, is the amount of water vapor available in the environment. When sk = s, the drop is in equilibrium with the environment. The drop will always try to achieve this equilibrium condition by growing (condensing ambient water vapor) or shrinking (evaporating water) until it reaches the size at which sk = s if it can! Another way to think about sk is that it is telling us something about the evaporation rate for each drop size and temperature. For the drop to be in equilibrium with the environment, the condensation rate of atmospheric water vapor must equal the evaporation rate of the drop. If sk < s, then net condensation will occur and the drop will grow. If sk > s, then net evaporation will occur and the drop will shrink.
Let's look at two cases. In the first case, the ambient supersaturation is always greater than the entire Koehler curve for a drop.
Interpretation of the figure:
In the second case, the ambient supersaturation intersects the Koehler curve:
Interpretation of the figure:
You can imagine all kinds of scenarios that can happen when there is a distribution of cloud condensation nuclei of different sizes and different amounts of solute. Drops with more solute have lower values for the critical supersaturation and therefore are likely to nucleate first because, in an updraft, the lower supersaturation is achieved before the greater supersaturation. So you can imagine the larger CCN taking up the water first and taking up so much water that the ambient supersaturation drops below the critical supersaturation for the smaller CCN. As a result, the larger CCN nucleate cloud drops while the smaller CCN turns into haze. The moral of the story? If you're a CCN, bigger is better!