Solutions to questions from chapter 8 in GEF4310 - Cloud Physics i.h.h.karset@geo.uio.no Problem 1 a) What is expressed by the equation below? Answer: The left side is the time rate of change of the mass of a particle (f.ex. liquid droplet), M w is the molar mass of water, while Φ v.sfc is the flux of water vapor at the surface of the particle. To calculate how fast the particle is growing because of the condensation of water vapor, we need to integrate over the whole surface area of the droplet. dm p = M w Φ v.sfc da b) What is the difficulties using this equation on a real problem? Answer: It is hard to find Φ v.sfc. We need to express it as a function of the ambient conditions (temperature, saturation, pressure, etc). c) What is Fick s law saying about Φ v.sfc? Answer: Fick s first law of diffusion is saying that the net flux of molecules is proportional to the concentration gradient, and that the flux is always directed opposite of the gradient (Φ v.sfc = D v n) Problem 2 We have three steps of droplet growth by condensation when following the Maxwell s theory. Which? Link them up to the equations below, explain the different factors and briefly explain the different steps. = 4πr d M w D v (n n eq ) = 4πr d M w D v n s (T ) l v dm p 4πr d k T M w T p T = src ( S S K ( ) ( )) lv Tp T 1 S K RT T Step 1 - Mass transport: Water vapor needs to be transported from the surroundings and towards the surface of the droplet. The mass growth rate of the droplet 1
because of this transport is described by the first equation. m d is the mass of the droplet, r d is the radius of the droplet, M w is the molecular mass of water, D v is the diffusivity, n is the concentration of water vapor far away from the droplet and n eq is the concentration of water vapor right next to the surface of the droplet when we have a steady-state vapor profile. Step 2 - Surface process: The droplets growth rate is more complicated because conditions at the surface affects it. We know from Calusius Clapeyron that the temperature of the droplet will effect how large the vapor pressure needs to be for the droplet to be in equilibrium (e eq ). The ideal gass law gives us a relationship between e eq and n eq, and since n eq is a part of the growth equation, the temperature at the surface of the droplet will impact the growt rate. After som calculations, we get the second equation. n s (T ) is the equilibrium vapor concentration over a flat pure water surface with the temperature T p, S is the saturation ratio far away from the droplet, S K is the equilibrium saturation ratio by the surface of the droplet and T is the temperature far away from the droplet. Step 3: Energy tranport: When vapor condenses at the surface, T p increases and becomes higher than T. Some of the energy released next to the droplet needs to be transported away if the droplet is going to continue the growth (since we know that larger T p requires larger e eq and larger n eq ). We get a flux of thermal energy directed away from the droplet. When reaching steady state, the temperature difference between the droplet, T p, and the ambient air far away from the droplet, T is given by the third equation. Problem 3 Correct or not? Explain why. a) During the growth of a droplet by condensation, the ambient supersaturations are higher than that right next to the droplet. Correct. We need this to create a mass flux from the surroundings towards the droplet. We get this since the droplet is continually taking up excess vapor. See also Figure 2. b) During the growth of a droplet by condensation, the steady-state flux of water vapor is large far away from the droplet and small right next to the droplet. Incorrect. The opposite is the case. See Figure 1 and this equation: dn dr = (n neq)r d. (Analogue: r 2 think of the flux of radiative energy from f.ex. the sun through a sphere right next to the sun and a sphere far away from the sun. Less energy goes through each m 2 far away from the sun) c) If the temperature of the droplet gets warmer, the mass rate of change due to condensation will decrease. Correct: See equation (8.11). Warmer droplet gives larger values of T p T. This makes smaller. (It is supposed to be a minus sign after S K (erreta)). 2
Figur 1: Figrue 8.3 from the book. Problem 4 Explain Figure 1. Answer: The figure shows three different steady-state vapor profiles when using Maxwell s theory of growth by condensation. The profiles (solid curves) shows how large the saturation ratios are at different distances away from the three different droplets surfaces (r d1, r d2 and r d4 ) when having steady-state. Steady-state here means that we don t get accumulation or sink in the vapor concentration anywhere between the surface and far away from it. The equation n(r) = n (n n eq ) r d r is the mathematical expression of the profiles we see in the figure (solid lines). We can see that the closer we get to the surface, the larger the gradient becomes. We can also see that the ambient saturation ratio is higher than that by the surface. Problem 5 Explain Figure 2. Answer: The figure shows how an increase in the temperature of the surface of the droplet due to condensation (the arrow from T to T p ) is affecting the vapor concentration we need at the surface of the droplet to be in equilibrium. The arrow from n s (T ) to n s (T p ) is showing this increase. The figure in the figure to the left is showing the Clasius Clapeyron relationship. Det curve down to the right is showing how the temperature profile from the surface (to the left) and away from the droplet (to the right) will look like after this heating by condensation. The solig curve up to the right is showing the same for the the vapor concentration, while the dashed curve right below it is showing the profile for the vapor concentration befor this heating. We can see that this profile is much less steep after the heating, making the flux of vapor weaker. Problem 6 Insert the missing words in theese important lines: 3
Figur 2: Figrue 8.4 from the book. The growth rate of a droplet growing from vapor is directly related to... Answer: the concentration of excess vapor concentration (n n eq ) The excess vapor concentration depends strongly on... Answer: the temperature of the droplet, T d. (Because n eq is a function of the temperature in the ideal gas law. The temperature of the droplet, T d depends on the... Answer: growth rate of the droplet. Problem 7 a) What does the following equation express? Explain the different factors. r d dr d = G (s s K) Answer: This is the linear growth law from Maxwell s theory. It expresses the growth rate of the radius, r d, of a cloud droplet due to condensation (or how it s decreasing due to evaporation). This equation also includes the surface processes and the impact of the condensation on the temperature and the flux of water vapor. All of this is included in the diffusivity factor, G. s s K is the difference in supersaturation between the surface of the droplet and the ambient air. b) How is the growth by condensation varying with the radius of the droplet? Answer: From the equation above, we can see that the larger droplets gives slower growth rate. c) We can se from the equation above that growth by condensation leads to a narrowing of the droplet spectra. How? Answer: The smaller droplets from the initial distribution will grow faster than the larger ones, but they won t catch up the droplet spectra will get more narrow in time. 4
d) What is the link between the equation above and the following equation? Rewrite the equation above so it becomes the equation below. = 4πr d ρ L G (s s K ) Answer: It is the same equation, but we have converted it from radius to mass with combining theese three formulas: m d = ρ L V d, V d = 4πr 3 d, We get that: Problem 8 = d ( 4 dr d = ( 4ρ L πr 2 d = dr d 3 ρ Lπr 3 d dr d ) G (s s K) r d ) G (s s K ) r d = 4ρ L πr d G (s s K ) What is the refinements of Maxwell s theory of growth by condensation (or shrinking by evaporation)? Just make a list. You don t have to explain all of then here, but say something about which sizes of cloud droplets the different points are relevant for. Answer: Assumes that the temperature difference between the droplet and the ambient air is small. This is not valid for evaporation of raindrops outside of the cloud Doesn t include ventilation effects. This is important for large droplets Doesn t include kinetic effects. This is important for small droplets. Doesn t include transient effects. Problem 9 a) When solving the growth rate equations in Problem 6, we often add a factor to make it more realistic. This factor is named f vent. What is it counting for? Answer: f vent is a ventilation coefficients, and counts for the ventilation effect we get when a liquid droplet sediment (fall) relative to the local air motions (because of density 5
differences). The ventilation makes the temperature fileds and the vapor fields distributed in a different way around the droplet than what would have been the case if the droplet was in rest relative the the surrounding air. b) How is f vent varying with the size of the droplet? List up some typical values of f vent. Answer: Since smaller droplets doesn t fall that fast compared to the surrounding air as larger droplets, the ventilation effect becomes larger for larger droplets. Below 60 µm, f vent is between 1 and 1.23. It increases with the size of the particle, and it is larger than 8 for raindrops (r d = 1 mm). c) What is the difference in the effect of ventilation inside a cloud compared to outside a cloud? Answer: We usually have growing droplets due to condensation inside the cloud and shrinking droplets due to evaporation outside the cloud. Both of the processes will speed up because of ventilation, so inside the cloud, the droplets will be larger when including this effect, but ouside they will be smaller. The effect will also be much more pronounced outside the cloud because of the larger raindrops compared to the smaller cloud droplets. Figur 3: Figure 8.6 from the book Problem 10 a) Briefly explain what we mean by kinetic effects. Answer: The Maxwell theory assumed that the vapor field and the temperature fields are continous throught the ambient air and all the way to the surface of the droplet. This is okay for larger droplets, but if we zoom in on smaller droplets, we can see that the fields are not continous. n is larger and T is lower right outside the droplet than what it should have been if we had a continous field and net balance between the evaporation and the condensation a the surface. It is molecular processes at the surface that causes 6
this jump in n and T right outside the droplet. We ve seen that the growth of droplets by condensation is dependent on both the temperature- and the vapor concentration profiles, so the kinetic effects will have an impact on the growth rate. a) What is the mass accomodation coefficient, α m and the thermal accomodation coefficient, α T? Answer: α m is the probability for a vapor molecule striking the liquid-gas interface to stay within the liquid (or the same as saying the probability of a vapor molecule that impinges the liquid-gas interface to incorporate into it). We often call α m for the condensation coefficient. α T is the probability for a gas molecule striking the liquid-gas interface to come into thermal equilibrium with the liquid before being diffusively reflected. The fact that theese rarely are 1, is the kinetic effects that we talked about in a). b) Explain Figure 8.6 and link it to α m and α T. Answer: Not all of the water vapor molecules that impinge onto the surface of the liquid droplet will not enter the liquid phase (incorporate into it). The Figure shows that the impingement flux of vapor is larger than the actual incorporation flux. We don t understand all of this process, incorporation flux impingement flux of vapor but we need to account for it. α m is equal to. We can also see that the energy flux out of the droplet can be different than that into it because the air molecules that strikes the interface can absorb some energy from the liquid since T surf usually is warmer than the surroundings due to condensational heating. The more energy the air molecules manage to bring out of the interface, the larger α T is. c) What is the typical values of α m and α T? Answer: From Figure 8.7, we get that α m is around 6%, while α T is around 70%, so we can say that while water vapor molecules that impinge the cloud droplets have a hard time of incorporate into it, air molecules that are hitting the droplet have less difficulties recieving energy from the droplet and transport it out into the ambient air. Problem 11 What is expressed in Figure 4? Answer: The figure shows how large the vapor concentration are at different distances away from the three different droplets surfaces with different sizes when having steady-state profiles. We can see that the concentrations are following increasing, smooth curves towards larger distances away from the droplets in the continuum regions, but that there is a discontinuity next to the surface (illustrated by this vapor jump) in the free-molecular regions. This is showing the kinetic effect of vapor molecules having a hard time incorporate into the cloud droplets. We can see that the vapor jump is decreasing as the droplet increases. 7
Figur 4: Figure 8.8 from the book Problem 12 a) What is expressed by the equation below? And what is R gas an R sfc? δn i n n eq 1 1 + αmcvr = d 4D v 1 1 + Rgas R sfc Answer: It is an expression of how the size of the jump in the water vapor concentration right next to the droplet because of the kinetic effect of vapor molecules having a hard time of incorporate into the cloud droplet when impinging it. When looking at how vapor molecules are transported from the ambient air and in the end incorporates into the droplet, there are two resistant mechanisms happening on the way. Firsty, the molecules needs to be transported from the ambient air and to the surface of the droplet: R gas = r d D v. Secondly, they need to incorporate into the droplet: R sfc = 4 α mc v. b) What is happening with δn i in the diffusion-limited case? Answer: When the droplet is large, r d is large, α m c v r d >> 1 and R gas will not be negligible compared to R surf. Solving the equation in a) with respect to δn i, we get that δn i = 4D v(n n eq ) α m c v r d. We can see that the larger the droplet gets, the smaller δn i becomes. c) How is Figure 5 related to the equation above? Answer: The figure shows how the growth of the droplet is affected by different values of α m (accomodation coefficient). From the equation above, we can see that the smaller α m gives larger δn i. We remember that small values of α m means that just a small part of the vapor 8
molecules that impinges the droplet will incorporate into it. This will decrease the growth of the droplet, as we see in Figure 5. d) The last mass growth law liquid droplets we end up with in chapter 8 is given here: = 4πr d ρ L G (s s K ) What does the equation express? Answer: It expresses the change in mass of a droplet because of condensation, but the kinetic effects are counted for (we can see that because of the G ). Except of the G, it is equal to the Maxwell theory in (8.17). e) How is G compared to G, and how does this affect the growth rate? G is smaller than G, making the droplets grow slower when including the kinetic effects. f) How is G changing with r d, and how will this affect the droplet spectra? Answer: When the droplet is growing larger, G G, so the larger the droplet is, the more equal to the Maxwell theory we get, and the less impact the kinetic effects have. Since the small droplets will grow slower due to kinetic effets, while theese effects are negligible for larger droplets, the kinetic effects broaden the droplet spectra. g How is the equation in d) related to Figure 6, how is the relation between r d and t, and what is the dotted lines pointing out? Answer: The figure is showing how the size of the droplet changes in time when it grow by condensation, but if we include kinetic effects. We get this if we rewrite the equation in d) from to dr d and solve it for r(t). Then we get that r d is proportional to the squareroot of t, r d t. This means that the droplets grows fast in the beginning when they are slow, but the larger the droplet gets, the slower they grow. The typical lifetime of a cloud droplet is around 1000 s (28 min). At this time, the droplets have only manage to grow to a size of r d = 20 µm. When we know that a typical raindrop is around 1000 µm, we know that there must be other mechanisms contributing to the growth. Problem 13 Very briefly explain what we mean by transient behavior. Answer: When deriving the Maxwell s equations, we assumed that the water vapor fields surrounding the droplet is in steady-state. This isn t always the case, for example under disturbances as turbulence, so it can affect the growth rate. When looking at the calculations on page 340-342 and Figure 8.14 in the book, we see that there are radpid adjustments of the vapor fields after a disturbance, so one can usually assume steady-state. Problem 14 Briefly explain the equation below. dm p = 4πCD v M w (n n p ) 9
Figur 5: Figure 8.11 from the book Figur 6: Figure 8.12 from the book Answer: This is the growth of ice crystals due to deposition of vapor in ice. It is written in the same way as the Maxwell equation for liquid droplets, but there are many assumptions because of the different shape of an ice crystal compared to that of a liquid droplet. The equation is derived using that the flow of water vapor towards the droplet can be described in the same way as electric current. That s why the C in the equation is named capacitance. C = Ce 4πε 0, where ε 0 is the permittiviy of free space and C e is the particle capacitance that is varying with the shape of the crystal (is it a needle? sphere? disk?). 10
Problem 15 a) What is expressed by the equation below, and how is it related to Figure 7? R B R P = dc/ da/ = dc da Answer: The equation expresses how the different faces of an ice crystal that is formed as a hexagonial prism is growing compared to each other. R B is the growth rate in the direction of the c-axis in Figure 7 (out of the basal face), while R P is the growth rate in the direction of the a-axis in Figure 7 (out of the prism face, out of the corners). b) Figure 8 shows that different temperature regimes favours growth in different faces of the ice crystal. Why is that so? Answer: The deposition coefficients, α m,prism and α m,basal, a measure of how efficient molecules enters the solid phase, depends on the temperature. This is seen in Figure 9. We will get at plate when the deposition coefficient for the prism phase is dominating, and we will get a column when the deposition coefficient for the basal phase is dominating. The reason for this temperature dependency is unclear. The growth is also depending on the differences in supersaturation between the different faces. c) For which temperatures are we most likely finding ice crystals shaped like columns? Answer: See Figure 8 and Figure 9. -3 C to -8 C and below -22. d) What do we mean by primary habits and secondary habits when talking about ice crystals? Answer: We can classify all crystals in either plates or columns. This is called primary habits. The crystals in each group can deviate from eachother by for example having sharper corners, beeing thicker or beeing longer. This is called secondary habits. e) Why do ice crystals shaped like a plate often grow faster at the corners than at the flat parts? Answer: Since the supersaturation vary around the crystal, and the values are often higher far away from the crystal, the corners can be located in a region with higher supersaturations. Problem 16 Explain Figure 10. Answer: The figure shows how large the temperature needs to be to melt ice crystals to liquid water under different relative humidities. We can see that the smaller RH becomes, the harder it is for the melting process to begin. The melting process starts when we get a liquid coating of the ice crystal, and energy is transfered towards the ice crystal through this interface. When the RH-values are low (f.ex. outside of a cloud), the liquid layer will evaporate to supply the air with moisture. Evaporation required energy, and the crystal-liquid-interface will cool. This supresses the melting. Inside the cloud 11
Figur 7: Figure 8.22 from the book where the RH-values are higher, the evaporation doesn t take place in the same order of magnitude, so it is easier for the melting process to begin. Problem 8.6.1 from the book Ignoring the curvature and solution effect, we have shown that the rate of growth of a droplet by vapor deposition can be expressed by r d dr d = G s. If a cloud sustains a supersaturation of s = 0.01 at a temperature of 4 C and pressure of 1000 hpa, calculate the time it takes a droplet to grow from 10 µm to 1000 µm. Discuss your results with respect to the formation of precipitation in cumulus and stratiform clouds. (Assume that s and T is constant during the process.) rd2 r d1 dr d r d = G s r d dr d = G s t 0 r 2 d2 r2 d1 = 2 G s t t = r2 d2 r2 d1 2 G s We have all the factors except of G given. Calculate G first using equation (8.18) in the 12
Figur 8: Figure 1.18 from the book Figur 9: Figure 8.23 from the book book. All the factors you need can be found in the book. Some of them is at page 184. G = = ( ρ L RT + ρ Ll v l v M wd ve s(t ) M wk T T 1 ) RT 1 S K 1000kgm 3 8.314Jmol 1 K 1 277.15K 18 10 3 kgmol 1 2 10 5 m 2 s 1 800Jm 1000kgm + 3 40.68 103 Jmol 1 3 18 10 3 kgmol 1 0.024Js 1 m 1 K 1 277.15K = 7.32 10 11 m 2 s 1 1 ( ) 40.68 10 3 Jmol 1 8.314Jmol 1 277.15K 1 (k T is the thermal conductivity of air, D v is the diffusivity of water vapor in air, l v is the latent heat of vaporization, M w is the molecular weight of water, R is the universal gas constant and ρ L is the density of liquid water.) 1 13
Figur 10: Figure 8.33 from the book Now we can calculate t: t = r2 d2 r2 d1 2 G s t = (10 3 ) 2 (10 5 ) 2 2 7.32 10 11 0.01 t = 682992 s 8 days The reason why we are asked to do the calculation up to 1000 µm is because this is the typical size of a rain droplet. We can see that it takes more than a week to from rain droplets by growth by condensation under normal atmospheric conditions. Rain droplets can form in less than one hour in the atmosphere, so there must be some other mechanisms involved. Problem 8.6.2 from the book Consider a raindrop falling from a tilted updraft through the side of a small Cororado cumulus cloud. Right before the drop exits the cloud, it is in steady-state growth in an environment with supersaturation s = 0.0001 and temperature 5 C. Assume that environmental conditions change instantaneously to that of the coller (3 C) and drier (s = 0.5) environmental air as the drop exits the cloud. Discuss what happens to the drop s vapor and thermal fluxes as it makes this trasition from supersaturated to subsaturated conditions. Answer: Inside the cloud, we have supersaturated conditions. This means that the droplet wil grow by condensation. When this happens, we will have a flux of water vapor ( Φ v ) from the surroundings towards the droplet. Since this will give condensational heating, there will be a thermal flux ( Φ T ) away from the droplet. The oppisite happens outside the cloud. Since the surroundings in the environmental air is subsaturated, the water vapor flux ( Φ v ) will be directed out from the droplet. The thermal flux ( Φ T ) will be directed towards the 14
droplet because it needs energy supply to break the bonds between the molecules during the evaporation. 15