Dg = 1 r Dp. g = 1 r p. = D ADr Atmospheric example: potential difference of the relative humidity

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1 Part III Cloud Physics 14 Evaporation How does evaporation work? Well, let s take what we know. The Second Law (Eq. 13.2) requires that for the system as a whole dg = d  m i ig i (T, p) < 0 dt dt The Gibbs-Duhem relationship (Eq. 12.6) gives us the relationship between pressure and temperature gradients and potential gradients DH m,p +V Dp H,V = mdg H,p If there is no energy barrier to initiating flows, we can constrain further for constant temperature (or enthalpy) mdg = V Dp or (Eq. 13.4) or as, a continuum expression (Eq. 13.5) Dg = 1 r Dp g = 1 r p Finally, pressure (or density) gradients drive flows (Eq ): m = D ADr t Dx T,p 14.1 Atmospheric example: potential difference of the relative humidity 1. Show that over a plane surface of pure water at constant temperature, where RH is the relative humidy, that the difference in the chemical potential µ between the vapor and liquid phase is Dµ = µ v (T ) µ l (T )=kt ln(rh/100) (14.1) where DG = mdg NDµ where N refers to the amount of matter (typically molecules) and Dµ is the amount of potential energy per molecule. The trick in the above is recognizing that, just above the liquid surface, vapor is in saturation (or equilibrium), such that its pressure is the same as the liquid water beneath it and Dp. Consequently, moving a vapor molecule from just above the liquid surface to the air above it is equivalent to moving it from the liquid phase to the vapor phase. 71

2 14.2 Atmospheric example: Why does water evaporate? 1. On the basis of the above, why does a puddle evaporate in dry conditions? Note that a system always work towards equilibrium. In this case a system always works towards a state where p v = p l. Hint: the chemical potential of vapor molecules just above a flat surface is what the chemical potential would be at saturation. Draw a box around the vapor and liquid system, and do an accounting of the total gibbs free energy in the box for the transfer of molecules, with their respective chemical potentials µ, from the liquid phase to the vapor phase Atmospheric example: How much water can air hold 1. It is often stated that the air can hold more water on a warm day. Comment 15 Pure droplet formation The basic question is what makes the existence of a droplet thermodynamically preferable to the existence only of water vapor. We have already derived an expression for the saturation vapor pressure over water. But what is the vapor pressure over a solution droplet? Atmospheric liquid clouds nucleate on tiny aerosol particles with dry diameters less that 1 µm diameter. These aerosol particles are the seeds of clouds. Without them, clouds could not exist in anything like the form we observe. Below we discuss some of the physics that is relevant to cloud formation Step 1 Gas molecules must collide and stick, to form a nucleus of a few molecules that is thermodynamically stable. It turns out that the stability of such a nucleus is favored by low temperatures (low particle energies). Thirteen molecules seems to be the magic number required to get things going. The physics of nucleation formation is beyond the level of this course. Step 2 Once, a nucleus is formed, a certain supersaturation is required to maintain the droplet in equilibrium with its environment, and a higher supersaturation to enable it to grow. We have shown that that Dµ = µ v µ l = 0ife = e sat (T ). This is strictly true only over a flat surface of water. Over a curved surface, it turns out that e > e sat (T ) for equilibrium. Qualitatively, the reason is that the total energy of the liquid surface includes not just the energy of water molecules themselves, but in addition the potential energy associated with the surface tension due to stretching the surface of the water into a curve. Consider that if the droplet were to burst, the surface tension would provide available energy to do dynamic work on the droplet. Therefore, at the same temperature, a curved droplet is at a higher potential with respect to water vapor than is a flat surface of water. This means that water vapor that ordinarily would be at higher potential than a flat surface water, and drive condensation, is now at a lower potential and is associated with evaporation. 72

3 Starting from the Gibbs-Duhem relationship (Eq. 12.6), expressed in terms of Dµ DH m,p +V Dp H,V = NDµ H,p We showed previously that, if there is no enthalpy barrier and DH = 0, then this can be expressed as Dg = Dp/r, or, per molecule (Eq. 14.1): Dµ = 1 n Dp = µ v µ v,l sat e (T ) = ktdlne = kt ln e sat (T ) where e sat (T ) is determined by the Clausius-Clapeyron Equation. If e > e sat then the current is from vapor to the droplet. If e = e sat, then the system is in equilibrium since Dµ = 0. We must now consider that there is an energy barrier to droplet formation due to surface tension forces. Suppose that enthalpy of surface tension is given by H = sa where A is the surface area of the droplet and s is the surface tension (units Newtons per meter or Joules per meter squared) which is about s = Nm 1 = Jm 2 Then the magnitude of the energy barrier produced, per molecule of droplet growth is dh/dt dn l /dt = dh dn l Defining the molecular enthalpy dh/dn l as Dḣ and n l = dn l /dv d as the molecular number density in a droplet of volume V d and radius r, Dḣ = dh dn = s da = 8prdr s n l dv d 4pr 2 = 2s dr n l n l r (15.1) Thus, from the Gibbs-Duhem expression (which we divide by N) we have a new expression for Dµ that includes the energy barrier from the surface tension for a droplet of radius r so Dµ tot = µ v µ sat l,v (T,r) Dḣ + 1 n Dp = Dµ tot Substituting our expressions for Dḣ (Eq. 15.1) and Dp, we now have 2s e + kt ln n l r e sat l,v (T ) = Dµ tot (15.2) Evaporation requires Dµ tot < 0 and µ v < µ l,v sat, allowing molecules to fall down, away from the liquid surface. Similarly, surface tension represents an energy barrier that always forces molecules away from the liquid surface. Thus, for Dµ tot to be positive, which would allow for condensation 73

4 onto a curved droplet, the vapor pressure e must be greater than it would be over a flat surface. In fact what is required is e ln e sat l,v (T ) > 2s n l ktr For a curved droplet to be in equilibrium with its environment, Dµ tot = 0 and we get the Kelvin Eq., which expresses the saturation vapor pressure over a curved surface with radius r 2s e sat l,v (T,r)=esat l,v (T )exp 1 (15.3) n l kt r or, perhaps more familiarly in terms of per mass e sat l,v (T,r)=esat l,v (T )exp 2s r l R v T 1 r (15.4) A couple of things to notice about this equation. First as r approaches infinity, the curvature effect on the saturation vapor pressure becomes insignificant. In other words, even if the droplet is a sphere, from a thermodynamic standpoint, it can basically be considered to be a flat surface. For atmospheric applications, the Kelvin effect can be neglected, and we can consider droplets to be essentially flat if they are larger than about 1 µm. Second, at the other extreme, for extremely small droplets, huge supersaturations are required for the droplet to be in equilibrium with its environment. For recently nucleated particles, say 0.1 µm across, the supersaturation at equilibrium is in excess of 100% (i.e. 200% relative humidity). We never get relative humidities this high in the atmosphere, and since any pure droplet forming from vapor must start from an embryo of a few molecules, clearly pure water droplets cannot form in the atmosphere. 16 Solution droplet formation 16.1 Kohler equation The Kelvin Equation (Eq. 15.3) has the form e sat l,v (T,r)=esat l,v (T )exp 2s n l rkt which was obtained by adding an energy barrier due to surface tension so that the total potential is Dµ tot = kt ln e e sat (T ) Thus we can interpret the Kelvin Eq. as being the equilibrium solution where Dµ tot = 0. We found that maintain equilibrium requires extraordinarily high values of supersaturation S = e/e sat 1 in order to compensate for the surface tension potential when the droplets are just nucleated and small. Somehow, we need to increase Dµ tot so that the vapor is at a higher or equal potential than the potential level of the droplet molecules µ v,l (T,r), so that being lazy, the droplet molecules are happiest to stay in the liquid phase. 74 2s n l r

5 T = 5 C 10 4 Supersaturation (%) mer R (µm) Figure 15.1: Dependence of the supersaturation of water vapor at equilibrium over a pure water droplet. The size of a 13 molecule nucleus is shown for reference. An effective way to do this is to add solute to the liquid. What this does effectively is to reduce the density of water molecules at the interface between the liquid and vapor phase, since the water molecules are separated by solute ions. Consequently the density of water vapor needs to be less in order for there to be equilibrium. Let s say we define a solution activity such that µ l (T,a w )=µ sat l (T,a w = 1)+kT lna w (16.1) where a w is bounded by 0 and 1. If a w < 1 then the chemical potential of liquid is depressed compared to what it would be at equilibrium over a flat surface of pure water. If a w = 1 then the chemical potential of the liquid is what we normally assume for pure water, and µ l (T,a w = 1)=µ sat l (T ) (16.2) Thus, since, right at the surface of the water, vapor is at equilibrium with the underlying liquid surface, and µ v sat (T,a w ) µ l (T,a w )=µ l sat (T )+kt lna w (16.3) we can alter our expression for Dµ from before to include the energy barrier representing the energy per molecule required to take the step upwards from the solution potential to the pure liquid potential, i.e. Dḣ = µ sat l (T ) µ l (T,a w ) µ sat v (T ) µ l (T,a w )= kt lna w (16.4) 75

6 The energy barrier that must be overcome to bring water vapor into equilibrium with a flat surface of pure water is Dµ = µ v sat µ v = kt lnrh. Thus, the water vapor is in equilibrium with the solution when the two energy barriers are equal and kt lnrh = kt lna w (16.5) where, a w corresponds to the relative humidity over a flat solution that brings the potential µ v into equilibrium with µ l (T,a w ). From the Gibbs-Duhem relation Dḣ + 1 n Dp = Dµ tot the combined potential difference per molecule between the vapor and the liquid phase, for a curved solution droplet, is now Dµ tot = µ v µ l (T,r,a w )=kt ln e e sat (T ) 2s n l r kt lna w (16.6) The second and third terms represent the sum of the energy barriers to diffusional flows Dḣ. If the activity of the solution can be lowered to a w < 1 then the last term is positive and this increases the energy barrier between the liquid and vapor phase Dµ. This makes existence of the liquid phase more stable and likely, despite the curvature effect. There is a bigger wall for the liquid molecules to jump over to get into the vapor phase, or a smaller wall required to initiate condensation. Our revised equilibrium expression then for the vapor pressure over a solution droplet is obtained by solving for Dµ tot = 0 This is known as the Kohler equation Activity of solution droplets apple 2s e sat (r,a w,t )=e sat (T )exp n l r + kt lna w 2s e sat (r,a w,t )=e sat (T )a w exp n l rkt e sat (r,a w,t )=e sat (T )a w exp For an ideal solution (usually true if very dilute) 2s r l R v Tr /kt (16.7) (16.8) a w = n 0 n + n 0 (16.9) where n 0 is the molecules of water and n molecules of solute. This is known as Raoult s Law. For dilute solutions a w = 1 n/n 0 (16.10) 76

7 But we must remember that salts dissociate, i.e. NaCl dissociates to Na + and Cl. We represent this using what is really a bit of a kluge, but popular nonetheless - the so called Van t Hoff factor i. n! in (16.11) where, for NaCl, and NH 4 HSO 4 (ammonium bisulfate) i = 2, and for (NH 4 ) 2 SO 4 (ammonium sulfate), i = 3. i is essentially the number of solute ions that the molecule dissolves into. The problem with this approach is that the number i stops meaning anything physical when one starts talking about organic non-ionic species, which nonetheless are soluble. What do we see from this? 1. The higher the concentration of the solute, the lower the equilibrium relative humidity 2. The more dissociation, the higher the solution effect. Effectively what is happening is that, while curvature makes it easier for molecules to escape a liquid droplet, dissolving a solute makes it harder for them to escape. Effectively there are fewer liquid molecules per unit surface area for them to escape. The Kohler curves are shown below. These show the relative humidity relative to saturation over a flat surface of pure liquid water that would be in equilibrium with a droplet of radius r containing dissolved solute. Effectively, it is e sat (r,a w,t )/e sat (T ) (NH 4 ) 2 SO 4 (r s =0.02 µm) (NH 4 ) 2 SO 4 (r s =0.03 µm) (NH 4 ) 2 SO 4 (r s =0.04 µm) (NH 4 ) 2 SO 4 (r s =0.06 µm) (NH 4 ) 2 SO 4 (r s =0.08 µm) (NH 4 ) 2 SO 4 (r s =0.1 µm) (NH 4 ) 2 SO 4 (r s =0.15 µm) (NH 4 ) 2 SO 4 (r s =0.3 µm) S r (µm) Figure 16.1: Kohler curves for ammonium sulfate. The location of peak saturation ratio S is at r, the critical radius. Below this size the solution effect dominates the Kelvin effect, and solution droplets are in equilibrium with their environment 77

8 (i.e Dµ = 0). If the particle grows spontaneously, it is then too large for the ambient equilibrium saturation for the droplet, so it shrinks to its original size. Likewise, if it shrinks it grow again to its equilibrium size. These are haze particles, which are in stable equilibrium. Above this size the haze particles are said to be activated. The Kelvin effect dominates and Dµ < 0 if the particles grow spontaneously. A particle that grows spontaneously will be have an equilibrium saturation lower than the ambient saturation, so it keeps growing spontaneously by vapor diffusion. The total Gibbs free energy of the system will continue to decrease. Once the droplets get big enough, the Kelvin effect stops mattering, and the droplet acts effectively like a plane surface of water. Therefore, the function of a solute in droplet formation is to lower the amount of chemical potential energy of the water vapor (i.e. supersaturation) required to get the droplet to the size r where it can start to grow spontaneously of its own accord. The supersaturations required are typically less than 1%, which is quite manageable in clouds. 17 Cloud droplet growth Once droplets are activated, how do they grow? There are two primary components to this problem. Growth by vapor diffusion, and growth by collision-coalescence. We showed that one of many legitimate forms for the non-equilibrium solution for flows from high to low potential is a current ~j that adds to the amount of mass m in a lower potential surface at constant T and p. From Eq ~j = m t T,p = A RT D p = AD r where, the sign convention here is for the direction to be from the high potential to the low potential, so that a negative density gradient in this direction corresponds with positive flow. Well, of course this could apply to the diffusional growth of a droplet, where the droplet is a low potential surface, and its super-saturated environment is a higher potential surface. Evaluated at the droplet surface, where the droplet has radius r, the mass increase m in condensed molecules, at the expense of vapor m v, and to surface area A = 4pr 2 follows mv t T,p = md t T,p = 4pr 2 D r x x=r where, the diffusivity D for the water vapor in air is about m 2 s 1. Also, here we ve switched the direction x so that it points radially away from the droplet. But, what is r/ x x=r? If we assume the ambient vapor field is in steady-state over time scales relevant to instantaneous flows, which is extremely close to being true, then from Eq dr/dt = 2 r = 0 If we assign x as the radial dimension, in spherical coordinates, we get 2 r x r 2x x = 0 78

9 which has the general solution r (x)=c 1 C 2 /x Applying the boundary conditions x!, r! r ; x! r, r! r r, we find the vapor field can be described by r (x)=r r x (r r r ) Thus, taking our flux solution, dm dt dm dt = 4pr 2 D r x x=r = 4prD (r v r vr ) (17.1) So rate of mass accumulation of droplets is proportional to their size. Note that the above equation is for a single droplet only, and that we would need to multiply the above equation by the droplet number concentration to get total mass density for a cloud. What stops droplets from growing indefinitely? Naturally, as droplets grow, they will deplete the available water from their surroundings, thereby reducing the vapor concentration gradient between the ambient air and the surface of the droplet. How fast do droplets grow in size? Starting from m = 4 3 pr lr 3 Using the ideal gas law for water vapor dm dt = 4pr l r 2 dr dt dr dt = D rr l (r v r vr ) dr dt = D rr l R v T (e e r ) dr dt = Dr v rr l e (e e r ) which, assuming e ' e sat (T ), such that the perturbation is not large, we can show that e e r e ' e e s e s S 1 = s where S is the saturation ratio and s is the supersaturation. So where r dr dt = G ls (17.2) G l = Dr v = Desat (T ) r l r l R v T We have made a number of assumptions in these derivations that we should be aware of. (17.3) 79

10 1. All water vapor molecules that hit the droplet stick. The reality may be that only 3% do. This matters mostly during initial growth of the droplet. 2. The temperature at the surface of the droplet is the same as the ambient air. In reality it is warmer due to the latent heat associated with condensation. It requires energy to transport this extra heat away from the droplet. This is an energy barrier that retards the rate of droplet growth 3. The droplet is stationary with respect to its environment. A droplet that falls as it grows is ventilated which changes the distribution of water vapor around the droplet and carries away latent heat. Apparently, however, this effect is very small, and can usually be neglected. If we integrate the growth rate equation we get q r (t)= r G lst (17.4) where r 0 = r, the activation radius if we are starting from a haze particle. A table of the amount of time it takes for a particle to reach a given size is given below. Note that droplets grow very Figure 17.1: The assumed supersaturation is 0.05%, p = 900 mb and T = 273K. rapidly at first, and as the droplets grow their initial size matters less and less. The same equations can be used to estimate the evaporation rate of droplets below cloud base It is easy to see from this table why the edges of clouds are so sharp. Only falling drops larger than about 0.1 mm will give any blurring to the cloud, and droplets must be significantly bigger in order to have any chance of hitting the ground, particularly if cloud base if very high (as it often is here in Utah in Summer). It is important to recognize that when a water drop evaporates, it does not completely disappear, but rather becomes a haze particle whose size is determined by the the Kohler equation according to the mass of solute it contains, and the ambient relative humidity (aka saturation ratio). 80

11 Figure 17.2: Droplet evaporation 18 Ice nucleation Liquid water can exist in clouds at temperatures below -30 C even. How then do we get ice in clouds? We discuss in class two basic formation mechanisms for ice Homogeneous nucleation Homogeneous nucleation is the freezing of a haze or water droplet at temperatures generally below 35 C. The physical principle is similar to that for homogeneous nucleation of water droplets, except that it occurs within in a water droplet rather than a supersaturated vapor field. If a droplet is sufficiently cold, by chance a few of the water droplet molecules will be moving slow enough to stick, and form an ice crystal embryo from which the entire droplet can turn to ice. Homogeneous nucleation appears to be the most important process responsible for producing large ice crystal concentrations in excess of 1 cm 3 in cirrus clouds, and is dominant particularly in the upper reaches of cloud with strong updrafts such as cumulonimbus. Recent studies show that a haze or water droplet freezes only when the saturation ratio with respect to ice is about 1.6 (compared to about for water clouds). Once a haze particle or water droplet freezes, it can grow directly by vapor diffusion. In general nucleation is the conversion of a liquid sphere to solid ice form. In this process a fraction of droplet will be frozen, with concentration N f, and a fraction will be unfrozen, with concentration dn u dn u = dn f dt dt The rate at which the conversion from liquid to solid takes place is given by the nucleation rate J (units number per volume per second) and the droplet volume and the number of unfrozen droplets remaining. dn u = N u V d J (T ) dt 81

12 Thus dn u = N u V d J (T )dt N u = N 0 exp( V d J (T )t) (18.1) The nucleation rate J (T ) itself is a thorny problem. However, studies by Koop in 2000 have shown a particularly interesting result. We know from Eq that, at equilibrium µ v = µ sat l + kt lna v w Similarly µ i = µ l sat + kt lna i w where a i w = a v w/s i, where S i = e s (T )/e si (T ). Note that the saturation vapor pressure over water is higher than that over ice, so S i can be greater than 1. Koop showed empirically that J = J a v w a i w (18.2) Well this is super useful, because at equilibrium, the value of a w should be the same as the ambient relative humidity (Eq. 16.5). Also, we know the saturation vapor pressure over ice and water. Thus, bingo, from the formula Koop provided (Eq. 18.2) and Eq. 18.1, we can calculate whether or not a haze aerosol should freeze to form an ice crystal at a given time t Heterogeneous nucleation Heterogeneous nucleation on the other hand involves an ice nucleus. An ice nucleus is usually composed of a material that has a similar crystalline structure to ice. By way of mimicking ice when such an aerosol particle comes in contact with a supercooled (<0 C) droplet, the droplet gets tricked into freezing. This process can occur at any temperature below freezing, but some ice nuclei require colder temperatures than others too initiate freezing. Perhaps the most important ice nucleus in the atmosphere is kaolinite (blackboard chalk) which is a common component of windblown dust and acts as an ice nucleus at an average temperature of -9 C. However concentrations of ice nuclei in the atmosphere are typically only about 1/litre, compared to at least 100/litre in clouds warmer than -35 C. It remains a puzzle exactly how to resolve this discrepancy. However, it appears that large concentrations of ice crystals are associated with clouds with initially high concentrations of liquid water droplets larger than about 20 µm. One possible mechanism for ice crystal multiplication that produces high crystal concentrations is the Riming-Splintering mechanism or Hallett-Mossop process. In this process an ice nucleus causes the outer shell of a droplet to freeze first, which then shatters to form more ice nuclei that can themselves initiate ice crystal production. However, this process appears to be most important between -3 and -8 C, and large concentrations of ice crystals have been observed at both colder and warmer temperatures. Although Hallett-Mossop is included in many mesoscale models, this is a big area of debate currently and far from resolved. 19 Ice crystal diffusional growth Generally air saturated with respect to water is supersaturated with respect to ice. This sets up a vapor pressure gradient such that in mixed-phased clouds, ice grows at the expense of water. 82

13 As before if the ice crystal is spherical, from Eq dm dt = 4prD (r v r vr ) where r vr represents the vapor pressure at the surface of the ice crystal. But as we know, ice crystals can be some truly crazy shapes. We can address this problem by drawing an analogy between the flux of vapor in and out of a crystal with charge leakage from a capacitor of the same size and shape. The capacitance C of a spherical capacitor is C = 4pre 0 (19.1) where e 0 = C/N/m 2 is the permittivity of free space. Forgetting about e 0, substituting Eq into Eq yields dm = CD (r v r vc ) (19.2) dt where r vc is the water vapor density at the surface. Following the same arguments as described for water droplets dm = CG i s i (19.3) dt where G i = Dr v ( ) and s i = e( ) e si e si Wallace and Hobbs outlines in Table 6.1 the types of growth that dominate. in some regimes it is the basal face that dominates and we get columnar shapes. In others, it is the prism face, and we get more plate like structures. In general, the greater the supersaturation with respect to ice, the more complicated the shape. Looking at diffusional growth more closely, consider Wulff s theorem, which is that for slow diffusional growth of ice crystals h b = s b ' 0.92 (19.4) h p s p where s is the surface tension of the ice crystal basal (b) or prism (p) face, and h is the height of the face. To understand this better consider the Kelvin equation we came up with, which is that e sat µ exp(ks) where k is a constant and ks is related to the energy barrier associated with surface tension that must be overcome before there is condensation. So the greater the surface tension the greater the saturation vapor pressure. This means that if we have a given vapor pressure e, then the speed of condensation will go as dm µ e e sat dt Therefore if the surface tension of the basal facet is is high compared to the prism face, then saturation vapor pressure is higher relative to the basal face, and the vapor pressure gradient e e sat is lower. This means that if s p is higher than s b, it follows that vapor will preferentially condense on the basal face: simply it is harder to condense on the prism face due to the higher surface tension there. It s a little counter-intuitive (just draw it to convince yourself), but the faster the 83

14 basal face grows, the large h p becomes, explaining Wulff s theorem. If s p > s b then h p > h b. Wulff s theorem explains the shapes of ice crystals well when supersaturations are very small and temperatures between -10 C and -22 C. The problem is that Wulff s theorem doesn t take us particularly far in explaining the wide variety of crystal aspect ratios that are observed at other temperatures and supersaturations. Growth is not necessarily very slow, extremely thin liquid layers can form on ice during the growth process, and ice crystals do not grow through simple vapor deposition but rather in steps and spiral formations. The true physics that controls ice crystal growth largely remains a mystery. 20 Observed cloud microstructures Here we discuss some the observed properties of warm (liquid water) clouds such as stratus and stratocumulus Vertical structure There are four basic parameters that are usually examined in measurements of clouds. 84

15 Figure 20.1: Number (left) and LWC (right) size distributions in clean stratocumulus and stratocumulus polluted by ships. 1. Size distributions (either, number, surface area, or volume). These look very similar to aerosol size distributions except the modes in the distributions are found at much larger sizes. The mode in the number distribution is typically between 10 and 25 µm diameter. The volume distribution is typically bimodal if the cloud is precipitating, with the second mode at somewhere between 100 and 1000 µm diameter. The precipiation mode has the largest sizes near cloud base, and the cloud mode (the small mode) has the largest sizes typically near cloud top. 2. The number concentration of droplets N (z)is usually nearly constant with increasing height. 3. The liquid water content LWC (z)usually increases linearly with height, starting near zero at 85

16 cloud base. 86

17 87 Figure 20.2: Vertical structure in stratus and stratocumulus off the coast of California

18 4. The volume mean particle size r v also increase with height, but starts at about 4 µm near cloud base, and increases less than linearly with height to somewhere between 5 and 15 µm diameter. Figure 20.3: Liquid water content profile in cumulus The vertical structure tends to reflect entrainment and precipitation which causes the LWC to drop off from its adiabatic value nearer to cloud top. 88

19 20.2 Horizontal structure Figure 20.4: Horizontal transect in cumulus The horizontal structure of clouds can be relatively homogenous, as in the case of stratus or fogs, or highly variable in stratocumulus or cumulus clouds. The degree of inhomogeneity appears to be linked to how turbulent the cloud is, and how it is precipitating. Observations show that the number concentration and liquid water content are much more highly variable horizontally than the droplet size Continental versus maritime clouds It is generally observed that continental clouds have smaller and more numerous droplets than maritime clouds. The degree to which this is the case appears to be closely linked to the concentrations of aerosols in the environment. 21 Observed rain and snow distributions Much like aerosol size distributions it turns out that rain and snow distributions can be closely parameterized by a simple exponential fit. The fit for rain drops is called the Marshall-Palmer distribution after the two scientists who first recognized this distribution based on a summer s observations in Ottawa Canada published in This distribution of droplets is characterized by the following equation dn(d) dd = N 0e LD 89

20 where D is the droplet diameter and N the concentration and L (units cm 1 ) is the slope of the rain distribution of a semilogarithmic plot. Marshall and Palmer showed that the slope can be related to the rainfall rate by L(R)=41R 0.21 where R is the rainfall rate and has units of mm hr 1. Strangely N 0 appears to be nearly independent of R and has an approximate value of N 0 = 0.08cm 4 (see Fig in Rogers and Yau) The equivalent equations for snow turn out to be L(R)=25.5R 0.48 N 0 = R 0.87 Figure 21.1: Snowflake shapes photographed by the muli-angle snowflake camera at Alta. Top row graupel, middle row rimed snow and bottom row aggregates. 90

21 1 Aggregates Rimed Graupel dn/dd max λ agg =0.69 mm 1 λ rim =1.19 mm 1 λ grp =1.99 mm D max (mm) Figure 21.2: Size distributions of snowflakes obtained with a Multi-Angle Snowflake Camera at Alta base. Concentrations are normalized. An example of snowflake shapes is shown in size distributions of snow measured at Alta is shown in Figure Associated size distributions are shown in Fig Note how riming produces more compact particles. 22 Atmospheric mixing This section introduces some elementary concepts associated with mixing and turbulence in the environment Mixing of conserved variables Studies of mixing of different airmasses often approach the problem by considering the conserved variables associated with each. Thus far we have discussed several of these 1. Conserved under unsaturated adiabatic processes q, h d, w 2. Conserved under saturated adiabatic processes 91

22 q e, h m, Q Note that w s and c are not conserved in general, and variables that are conserved under dry processes are not conserved under moist processes. Now suppose that we have two airmasses that meet and mix, without interaction with the surrounding environment. Such mixing is a fundamental aspect of for example heat transport from the equator to the poles, or temperature and humidity gradients along fronts, or entrainment of dry air into clouds. How do we determine the state variables associated with the mixed parcels? By way of illustration if we take mass fraction f from one air parcel A and mass fraction (1 f ) from a second parcel B and there is no saturation during the mixing process then the dry static energy and mixing ratio associated with the new parcel is h d = fh da +(1 w = fw A +(1 f )h db f )w B Since h d = c p T + gz, if one knows the height of the new parcel, one now knows the temperature and humidity of the new parcel Adiabatic Mixing without Condensation We derive here the basic equations dealing with adiabatic mixing without condensation. Assumptions The two parcels with masses m 1 and m 2 mix adiabatically The two parcels are at the same level (d (gz)=0) (e.g. mixing across a front). Since the two parcel are at the same level we are mixing only enthalpy and water vapor. Therefore, The change in enthalpy is then m 1 h 1 + m 2 h 2 =(m 1 + m 2 )h DH = 0 = m 1 (h h 1 )+m 2 (h h 2 ) so the new temperature is m 1 Dh 1 + m 2 Dh 2 = 0 m 1 c p (T T 1 )+m 2 c p (T T 2 )=0 T = m 1T 1 + m 2 T 2 m 1 + m 2 Similarly w = m 1w 1 + m 2 w 2 m 1 + m 2 92

23 e = m 1e 1 + m 2 e 2 m 1 + m 2 q = m 1q 1 + m 2 q 2 m 1 + m 2 Example Figure 22.1: Two subsaturated parcels combining to produce a saturated parcel. Note that both e and T mix linearly, but that clearly RH does not. Two parcels of air mix thoroughly across a cold front at 1000 mb. Parcel one has a temperature of 23.8 C and w = 16g/kg, and parcel two has a temperature of T = 12.4C and a mixing ratio of 5 g/kg. 1. If both parcels of air mix equally, what is the final temperature and mixing ratio of the combined air masses? Since the parcels mix equally, the new temperature and mixing ratio is simply the average temperature and mixing ratio, which are 18.1 and 10.5 g/kg. 2. What is the initial RH of each parcel and final RH? It works out that e s (23.8 )=29.5mb 93

24 and that the saturated mixing ratio is So the relative humidities are e s (12.4 )=14.4mb e s (18.1 )=20.7mb w s = e s p w s (23.8 )= w s (12.4 )=0.009 w s (18.1 )= RH (23.8 )=87% RH (12.4 )=56% RH (18.1 )=82% So unlike T and w, RH does not mix linearly! This is an important consideration since it leads to the observation that two subsaturated parcels of air can combine to produce a cloud Figure 22.2: Sea-smoke A most striking example of this sort of this is when a cold dry air mass moves over a warm moist air or surface to produce sea-smoke. Another example is when warm moist exhaust from an airplane mixes with cold dry air to produce a contrail. 94

25 23 Atmospheric turbulence 23.1 Characteristics Examples of Turbulence Cumulus clouds Jet streams Dust storms Characteristics irregularity diffusivity - no spreading no turbulence vorticity dissipative due to viscous losses (waves are non-dissipative although they can be dispersive) characteristic of fluid flows Associated with large values of the Reynolds number Re where Re = ul/n, where L is the characteristic length scale of an object moving with speed u through a viscous medium with kinematic viscosity n. Sources shear in a mean flow buoyancy The transition from laminar to turbulent flow is one of the more poorly understood aspects of physics. On his death bed, the famous fluid dynamicist Horace Lamb said, ""I am an old man now, and when I die and go to heaven there are two matters on which I hope for enlightenment. One is quantum electrodynamics, and the other is the turbulent motion of fluids. And about the former I am rather optimistic" 23.2 Length Scales in Turbulent Flows Turbulent flows have a wide range of length scales covering many orders of magnitude in size ranging from the dimensions of the flow itself at the high end, to the length scale associated with the diffusive action of molecular viscosity at the low end. 95

26 Laminar Boundary Layers What are the relevant scales for laminar, low-shear flow? The Navier-Stokes equations for steady flow are u du dx = 1 dp r dx + n 2 u x 2 (23.1) The first term is inertia, the second the pressure gradient force, and there is a non-linear third term, the viscous term with n = 0.15 cm 2 s 1, which is needed to create turbulence. For viscosity to be sufficiently large to damp turbulence the viscous and inertia terms have to be of the same magnitude. Doing scale analysis we would require U 2 L nu L 2 Dividing the first term by the second, we obtain the Reynolds number UL n Re (23.2) If the Reynolds number is of order unity than turbulence is damped and flow is laminar. But Reynolds numbers typical of the atmosphere are typically on the order of a few thousand, so the atmosphere tends to be turbulent. In clouds the source of energy is mostly latent heat release, and this creates turbulent motions while sending air upwards. With respect to the turbulence, it is common to talk about an energy dissipation rate e with units of energy per mass per second (m 2 /s 3 ). It represents the rate at which energy is transfered from large-scale motion to small scales. Eddies start big and then become progressively smaller. But how small before the flows become laminar because viscosity takes over? Combining the rate at which the smallest scales receive energy with the viscosity, using dimensional relationships we can derive length, time, and velocity scales for the smallest scales as follows h = n 3 /e 1/4 (23.3) t =(n/e) 1/2 (23.4) v =(ne) 1/4 (23.5) These are called the Komolgorov microscales of length, time, and velocity. The corresponding Reynolds number formed with these numbers is equal to one hv/n = 1 Small-scale motion is highly viscous, and that the viscous dissipation of turbulence as heat adjusts itself to the energy supply e by adjusting length scales h. Can we estimate what the supply rate of energy e is to small scales? Well the energy comes from large scale turbulence. This energy is proportional to u 2. It seems appropriate to assume that the time scale T associated with dissipation of the energy is then u/l which is roughly one overturning of the eddy. Therefore the rate of energy dissipation is of order u 2 /T = u 2 u/l. e u 3 /l (23.6) 96

27 Figure 23.1: Analysis of turbulence data 97

28 Let s say you measure a time series of wind speed. By doing a fourier transform on this time series you can obtain a power spectrum of the data. It works out that for there to be a cascade of energy from large scales to small scales requires that the slope of this energy spectrum be - 5/3. At one end of the spectrum are the eddies of speed u with length scale l. At the other end is the Kolmogorov microscale with speed and length scales v and h. The rate at which energy is transferred between the two is e, and this is also the rate at which large scale turbulence gets converted to heat. The power spectrum continues leftward to larger length scales, but these tends to be associated with gravity waves first, and synoptic scale waves second Power dissipation by a cumulus cloud Figure 23.2: A cumulus cloud Estimate the energy dissipation rate in a cumulus cloud, both per unit mass and for the entire cloud. Base your estimates on velocity and length scales typical of cumulus clouds. Compute the total dissipation rate in kilowatts. Also estimate the Kolmogorov microscale h. Use r = 1.25kg/m 3 and n = m 2 /s. We can estimate the dissipation rate using e u 3 /l In a cumulus cloud a typical turret (arguably the length scale of the largest eddies) is about 250 m. The updraft velocity is about 1 m/s. This gives values of e m 2 /s 3 for the energy dissipation rate per unit mass. The energy dissipation rate of the entire cloud, assuming fairly suitable dimensions of 1 km 3 and r = 1.25 kg/m 3 is m e = = Js 1 = 5000KW 98

29 (How does this compare to energy production associated with latent heat release during ascent?). The Kolmogorov microscale is h = n 3 /e 1/4 = ! 1/ = = 1mm So energy gets converted to heat in cloud at spatial scales on the order of 1 mm. Pretty small! Typical values of e range from 10 4 to 10 2 m 2 /s 3 in convective clouds to 10 5 to 10 4 m 2 /s 3 in stratiform clouds. References Tennekes, H. and Lumley, J. L., A First Course in Turbulence, MIT Press, 1972, pp 300. Seinfeld, J. H., and Pandis, S. N. Atmospheric Chemistry and Physics, Wiley Interscience, pp