Solutions to questions from chapter 11 in GEF Cloud Physics

Transcription

1 Solutions to questions from chapter 11 in GEF Cloud Physics i.h.h.karset@geo.uio.no Problem 1 a Draw a sketch of how the radius of a cloud droplet is changing in time. Just for radius up to 50 µm. Answer: See Figure 1 Figure 11.3 in the book. b Why are the curve different during the time where is the dominant growth mechanism, compare to the time where dominates? Answer: When the droplets grow by, the growth rate is given by eq in the book. Solving it for rt, we ve seen that r t See f.ex. Figure 8.12 in the book. This is because the smaller droplets change their radius fast when some extra mass is added during, but the change in radius for the larger droplets are not that big when extra mass is added because the surface area where can happen is much larger compared to the volume for a small droplet compared to that for a larger droplet. During, the large collector drop becomes more and more efficient in when it grows larger, so the growth will be exponential, at least in the beginning. c What is happening in the size range around 20 µm? Answer: The is not efficient in increasing the radius fast when the droplet is approaching 20 µm. The is not efficinet yet for these sizes either, so other mechanisms needs to be contributing for the growth in this region. Problem 2 A typical cloud droplet has a radius of 10 µm. A typical rain drop has a radius of 1 mm. How many cloud droplets does a raindrop contain? Answer: We have to look at the 1

2 Figur 1: Figure 11.3 from the book. relative masses to see how many cloud droplets we get in a raindrop: m r m c r3 r r 3 c 4 3 πr3 rρ L 4 3 πr3 c ρ L We can see that we need the mass from 1 million cloud droplets to create a raindrop. Problem 3 a How can entrainment contribute to a broadening of the droplet spectra and thus increase the probability for some drops to grow big enough to be effective collector drops? Answer: If the mixing is inhomogenous, some of the droplets will evporate completly and some of the droplets will be unaffected by the mixing. The excess water vapor due to the evaporation can be used for the remaining droplets to grow further by. The smaller the cloud droplet number concentration is N d, the less competition it is for the excess water vapor, and the bigger the droplets can grow. This is seen by the expression for the quasi-stationary supersaturation, s QS Aw r d N d. 2

3 Figur 2: Figure 11.5 from the book. supression of the growth and delay the start of the process? Answer: If the mixing is homogenous, all droplets will experience some evaporation, which will set back the growth. b Which of the two cases is linked to the evolution of the cloud droplet number concentration seen in Figure 2? Answer: The inhomogeneous entrainment event, a1. Problem 4 a What is the difference between observed time for rain to form and calculated mean time for rain to form? Answer: Observed time is much shorter. b How is the theory of stochastic explaining these differences? Link it up to Figure 3. Answer: The process are stochastic, so it is a possibility for it to start before the droplets reach a certain value, but it s also possible that it starts later. Some droplets are lucky and collide and coalesce early, giving rise to larger droplets earlier than the mean time. We know, based on Problem 2, that we only need one larger droplet per 10 6 smaller droplets to develope raindrops, so this could be the explaination why we observe that rain forms faster than calculated by using the mean value for when will start. Figure 3 illustrates this. c Figure 3 shows that the growth by following the theory of stochastic is not continous in the beginning. Why is that so? Answer: The events are based on probabilities, so they will only occur at the times when the probabilities of finding another drops to collide and coalesce with are with them. The smaller the droplets, the smaller the probabilities, so we won t get a continous curve in the beginning because there will be times where no will happen. 3

4 Figur 3: Figure 11.6 from the book. Problem 5 a What is expressed by the equation below? Explain the equation. Answer: It s the stochastic theory describing the evolution of the drop spectrum due to collision-coalescence. nm, t is the concentration of cloud droplets with a mass within the interval m + dm at time t. This concentation can decrease when droplets with this size is growing bigger, and it can increase when smaller droplets merge and form a droplet with a mass in this interval. The first integral expresses the source term. The factor of 1 2 is there because it is two droplets with masses m x and m m x that is contributing to each raise in the concentration within this interval, but we can t count them twice, because we only get one bigger droplet with mass m out of the collision. The K-factors are called kernel, the effective volume swept out by the collector drop in unit time see Figure 11.4 in the book. nm, t t 1 2 m 0 Km x, m m x nm x nm m x dm x nm 0 Km, m x nm x dm x b How is the equation above linked to Figure 4? Explain the Figure. Answer: The Figure shows how the droplet spectra will be affected if one allow stochastic given in the equation above compared to continuous. Continuous only let large droplets collect very small droplets at one specific time. The stochastic process allows smaller droplets to collide and merge, and it is probability functions that descides whether or not that will happen. We can see that stochastic contributes to a broadening of the droplet spectra, where some droplets gets bigger than the others after just a short time period. c Is Figure 5 illustrating stochastic or continuous growth by? Answer: Stochastic. We can see this because of the time. Cloud droplets are transformed into raindrops within 30 minutes. That s not happening with continuous. d The Bowen model can be used to calculate the rain development in warm clouds. Is it based on stochastic or continuous growth by? Answer: continuous 4

5 Figur 4: Figure 11.8 from the book. Figur 5: Figure 11.9 from the book. Figur 6: Figure from the book. 5

6 Figur 7: Figure from the book. Problem 6 a Explain Figure 6. Answer: Cloud droplets are lifted above the cloud base. When they have grown large, the updraft isn t strong enough to compensate for the downward gravitational force, so they will fall down. The droplets grow because of of smaller cloud droplets. b Why will cloud droplets within a cloud with large updrafts become larger than cloud droplets in other clouds? Answer: the larger updraft, the higher up in the cloud the droplets will reach before the downward forces are strong wnough to compensate for the updraft. This will lead to a longer pathway for the droplets to travel on their way down, which means that they have time to collect many smaller droplets before they fall out of the cloud. c Figure 6 and Figure 7 are illustrating growth of cloud droplets to raindrops according to the Bowen model. Based on what you can see in the figures, what do you think is the main reason why this model isn t complete for describing the developement of rain from warm clouds? Answer: Rain is produced within minutes. Figure 6 shows that it takes more than one hour. Problem 7 a The expressions below are used to describe the size distribution of cloud droplets in a cloud. What is the name of the different distributions? Answer: Log-normal distribution and the gamma distribution. nd nd N tot N tot e 2π D ln σg ln D ln Dm 2 2 ln 2 σg 1 D ν 1 1 Γν e D Dn D n D n 6

7 Figur 8: Sketch of the size distribution functions used for cloud droplets. b What is common for the two distributions? Draw a sketch of how they usually look like with normal values of the different parameters. Answer: They have a long tail against larger values of D, and a fall-off in concentrations at small sizes. Se Figure 8 c What is ν? And how will the look of the graph for the distribution function change when ν increases? Answer: ν is the shape parameter, telling something about the shape of the curve. If this value is 1, the curve is just exponential decaying. When ν is larger than 1, the curve look like the one in Figure 8. The larger ν gets, the more equal to the normal distribution the graph will look like. d What is D n, and how will the look of the graph for the distribution function change when D n increases? Answer: D n is a scaling parameter. The larger it gets, the higher the peak of the curve will be, and it will also be shifted towards lower values. It is closely related to the mean diameter. e In cloud physics, it s common to write the Gamma ditribution like this: nd n 0 D β e ΛD. What do we call this distribution when β 0, n m 3 mm 1 and Λ 41R 0.21 And what is β and R? Answer: β is closely related to the shape parameter in c, β ν 1. R is the rain rate. This modified gamma distribution is called the Marshall-Palmer distribution. f According to the Marshall-Palmer distribution, how is nd varying with D, And how is this seen in Figure 9? Answer. nd is decreasing exponentially when D is increasing. This is seen by the dot-dashed straigth line in Figure 9. g How and why is the observed drop-size distribution in clouds varying from the Marshall-Palmer distribution? Answer: Not so many small droplets because they are collected by the bigger ones. Not so many very large drops because the will break up. See the solid, black, bold line in Figure 9. 7

8 Figur 9: Figure from the book. Problem 8 a What is the first aerosol indirect effect, And what is another name for it? Answer: When the aerosol concentration is large, the water inside a cloud is spread among many droplets instead of just a few. Even though the droplets are smaller than in cleaner air, the total total surface area of the cloud droplets is larger. This will reflect more of the incoming solar radiation. It s also called the Twomey effect. b What is the second aerosol indirect effect, and why isn t observations always showing this? Answer: When clouds get polluted by more aerosols and the droplets are small, precipitation can be supressed because the droplets are too small to initiate the precipitation proscess. This can give thicker and more longlived clouds that can reflect more of the incoming solar radiation. We don t always observe this because of many different reasons. Examples: 1 The aerosols can be big giant aerosols that can initiate precipitation. 2 Supressed precipitation can reduse the stability, increase the turbulens, increase the entrainment and thus reduce the thickness of the clouds. Problem from the book Calculate the ratio of the rate of continuous to vapor depositional growth in a cloud with a temperature of 4 C at a pressure of 1000 hpa, supersaturation s 0.01 and liquid water concentration ω L 0.2 gm 3 and ω L 1 gm 3 a a 60 µm diameter drop b a 140 µm diameter drop Discuss your results. Answer: We are going to calculate this relationship: EckωL 4ρ L r d 1 r d Gs 8

9 Assume E c 0.5, k s 1 and ρ L 1000 kgm 3. Use equation 8.18 in the book to calculate G same way as we ve done earlier weeks... and get that G m 2 s 1. Insert all the values for all the cases, and get that: D d 60 µm and ω L 0.2 gm 3 : 21 µmh 1 90 µmh 1 D d 60 µm and ω L 1 gm 3 : 108 µmh 1 90 µmh 1 D d 140 µm and ω L 0.2 gm 3 : D d 140 µm and ω L 1 gm 3 : 50.4 µmh µmh µmh µmh 1 We can see that the larger the droplet gets, the more inefficient the is to increase the radius because of the smaller surface area compared to the volume. We also see that the larger the droplet gets, the more efficient the process becomes because they sweep over a larger area when falling, and can collect many more droplets. We see that the liquid water concentration plays a role. The larger ω L is, the more efficient the process becomes there s many more smaller droplets to collect when the LWC is large. 9