2 D. Terminal velocity can be solved for by equating Fd and Fg Fg = 1/6πd 3 g ρ LIQ = 1/8 Cd π d 2 ρ air u

Transcription

1 ecture 8 Collection Growth of iquid Hydrometeors. I. Terminal velocity The terminal velocity of an object is the maximum speed an object will accelerate to by gravity. At terminal velocity, a balance of opposing forces is attained. A. The balance of forces. Terminal velocity is a balance of gravity and drag forces due. 1. Gravity accelerates particles downward. 2. Drag forces are a result of resistance of air molecules to the motion of some object. B. Gravity Force (Fg) Fg = V g (ρ IQ - ρ air ) V is the volume (1/6πd 3 for a sphere) g is gravity ρ IQ is the density of liquid drop ρ air is the density of air ρ IQ >> ρ air 1. This relation is approximated as (for a sphere) Fg = V g ρ IQ = (mg) Fg = 1/6πd 3 g ρ IQ where d is the diameter C. Drag force (Fd) Fd = 1/2 Cd A ρ air u 2 Cd is the drag coefficient which is a function of shape, size, characteristics of surface, Reynolds number (See Fig. 3) A is the cross sectional area normal to direction of fall For a sphere, A = π/4 d 2 u is terminal velocity 1. The drag force relation can be written as (for a sphere) Fd = 1/2 Cd A ρ air u 2 = 1/8 Cd π d 2 ρ air u 2 D. Terminal velocity can be solved for by equating Fd and Fg Fg = 1/6πd 3 g ρ IQ = 1/8 Cd π d 2 ρ air u 2 = Fd 1. Which results in... u = [(4 d ρ IQ g) / (3 Cd ρ air )] 1/2 2. or, u = [(4 ρ IQ g) / (3 Cd ρ o,air )] 1/2 d 1/2 [ρ o,air /ρ air )] 1/2 = c D 1/2 [ρ o,air /ρ air )] 1/2 where ρ o,air is a reference density of air E. Is the relation derived in D. a reasonable approximation for water droplets of all sizes? [ u = c D 1/2 [ρ o,air /ρ air )] 1/2 ] 1. This equation is valid for spheres. Are droplets always spheres? See Fig. 4 on attached page. 2. Gunn and Kinzer plotted observed terminal velocity as a function of size and found the following power law relation: u = c D 1/2 with c = , and d = 0.8 a. This equation is valid for 100µm < D < 6 mm 3. Other approximations: a. D < 80µm 1

2 u = c D p = c D 2 with c = [(ρ IQ g) / (18 ν ρ air )], p = 2 [SEE F BEOW] b. 80 < D < 1200 µm u = c D p = c D with C = 4x10-3, p = 1 4. Plots terminal velocity as a function of size are shown in Fig. 5. F. The drag force is often written as. Fd = 3 π ν ρ IQ d u [Cd N RE / 24] 1. From this, u = D 2 ρ IQ g / [ 18 ν ρ air {Cd N RE / 24} ] 2. Why do this? a. [Cd N RE / 24] --> 1 for small Reynolds number (d<80µm) b. As a result u = c D p = c D 2 with c = [ρ IQ g) / (18 ν ρ air )], p = 2 II. Conditions eading to Collectional Growth A question for much research in the past/present was how could condensation and coalescence (collectional growth) result in rain drop sized hydrometeors (1 mm) in times of 20 minutes or so from first cloud to first rain. This time is not even considered well known. Growth begins by diffusion of vapor on an initial spectrum of nuclei. Further diffusional growth leads to spectrum broadening. Until a few droplet sizes reach 20µm or so, however, growth by coalescence is very small because of the small differences in terminal velocities of droplets and the small probability that they will coalesce. Once droplets are 30 or 40µm in radius, growth by collection becomes more dominant. Further growth broadens the spectrum, increases the differences in terminal velocities of droplets, and accelerates collisional growth. III. Some Basic Ideas about Collectional Growth. A. Collisions occur due to 1. Gravitational forces a. Drops of different sizes fall at different speeds 2. Electrical forces a. Oppositely charged particles may be attracted to each other 3. Aerodynamical forces a. Turbulence, etc. may enhance growth B. In general, gravitational effects dominate over electrical and aerodynamical forces. C. We need to be concerned with 1) whether droplets will collide, and 2) whether droplets will coalesce (stick). When droplets collide, the following can happen: 1. coalescence 2. rebound (collide and bounce apart) 3. coalesce and then separate with original sizes intact 4. coalesce and then separate with different sizes, possibly producing additional droplets. D. The important variables in collisional growth are: 1. Sizes of droplets involved 2. Fall velocities 2

3 3. trajectories of droplets 4. Number of collisions 5. Number of sticking collisions (i.e., actual droplets that coalesce) 6. Electrical and turbulent effects E. Collisional growth goes by many names 1. Collisional growth 2. Accretional growth 3. Coalescence growth 4. Riming growth 5. Aggregational growth IV. The Basic Collection Equation In general, the collectional growth problem is a many-body problem. However, arguments can be made to consider collectional growth in terms of a simplified two-body problem. A. Two-body collection 1. Justification a. If we have 1000 droplets per cm 3, and the average droplet is 10µm in diameter, then they have an average spacing of about 100 droplet diameters (1000µm): It can be said that droplets are relatively sparse in a cloud. 2. Assumptions a. Assume we have two spherical droplets of different sizes i. Shape effects can be considered to be of secondary importance for our basic problem at this point. b. One droplet is larger than the other, therefore it is more massive (both droplets have the same density) and the larger falls faster. c. The droplets are assumed to be falling at their terminal velocity d. Droplets are initial widely separated in the vertical e. Droplets are vertically aligned f. The air and internal fluid motions are calm 3. The basic physical model a. We seek to find the effective cross sectional area for the two droplets b. The Fig. 6 shows a graphical representation of the physical model. i. R is the radius of the large drop, R S is the radius of the small drop. ii. The geometric sweep out area for the two droplets is π (R + R S ) 2 iii. The geometric volume / per second swept out by this two body system is π (R + R S ) 2 (v - v S ) where v and v S are the fall velocities of the smaller and larger droplets. 3

4 c. The volume increase of the larger droplet is given as dv /dt = 4/3 π R S 3 N S π (R + R S ) 2 (v - v S ) d. The mass increase of the larger droplet is then dm /dt = ρ w 4/3 π R S 3 N S π (R + R S ) 2 (v - v S ) e. And finally, a radius increase rate is 2 ) NS π (R + R S ) 2 (v - v S ) using M = ρ w V = ρ w π 4/3 R 3 and dm /dt = ρ w 4 π R 2 dr /dt V. Collision Efficiencies Whether or not two droplets will collide is dependent on the relative importance of the aerodynamical and inertial forces associated with the flow around each droplet and how the flows interact when the two droplets approach each other. A. The grazing trajectory is the farthest distance the smaller droplet can be from the fall line of the larger droplet and just touch the larger droplet. 1. The grazing trajectory radius is denoted by R in Fig All small droplets (relative to the larger droplet) that lie within radius R of the fall line of the larger droplet can be collected. All droplets that lie outside radius R are not collected (at least in calm / non-turbulent conditions). B. The simplest collision efficiency E is defined as: E(R,R S ) = πr 2 / [π(r + R S ) 2 ] = R 2 / [(R + R S ) 2 ] where R is the grazing trajectory radius as defined before. 1. The linear collection efficiency can be thought of in two ways. a. Fraction of droplets with radius R S in the path of the collector droplet of radius R that are actually collide with it b. Probability that a small droplet randomly located in path of collector droplet will collide with it C. The collision efficiency is often defined as E(R,R S ) = R 2 / R 2 1. This can be derived by assuming R >> R S E(R,R S ) = πr 2 / [π(r + R S ) 2 ] = R 2 / [(R + R S ) 2 ] = R 2 / [R 2 ] D. Typical values of collision efficiencies are shown on the attached figures. 1. For R 2 / R 2 << 1, E is <<1. a. Physically, the small droplets have almost no inertia, and are easily deflected by aerodynamic forces around a large collector drop. 2. As R 2 / R 2 -->0.6, E approaches its maximum 3. For R 2 / R 2 --> 1.0, E can begin to decrease a. Two droplets, with nearly the same radius, approach each other very slowly as their terminal velocities are similar. The droplets can be deflected around each other. b. Wake effects can make values of E > 1. 4

5 i. These have been experimentally and numerically verified 4. In general, E is an increasing function of R and R S. For R > 100µm, E is primarily dependent on R S. E. Coalescence 1. Collision efficiency does not take into account the probability of coalescence (sticking). Theoretical studies generally provide collision efficiencies, while observation studies provide collection efficiencies (coalescence times collision). 2. When droplets collide, the following can happen: a. coalescence b. rebound (collide and bounce apart) c. coalesce and then separate with original sizes intact d. coalesce and then separate with different sizes, with the possibly of producing additional droplets. 3. Typically we assume that coalescence efficiencies are unity. Electrical effects often play a significant role in determining coalescence efficiencies. 4. Studies show that coalescence efficiencies can be a function of the Weber Number (see your book and Fig. 9). Note that the Weber Number is the ratio of dynamic pressure to pressure across a curved interface: It is given as N WE = (ρ w R / σ l/a ) (v - v S ) 2 a. Figs. 10 show more plots of coalescence efficiencies 5. Fig. 11 show a plot of collection efficiency (coalescence times collision efficiency) 6. Fig. 12 shows the effect of impact angle of a collision vs. relative terminal velocity and the resultant collisional behavior. As noted before, droplets can rebound, coalesce, or cause disruption. E. The growth equations can now be written as the following shows: 1. The volume increase of the larger droplet is given as dv /dt = 4/3 π R S 2. The mass increase of the larger droplet is then dm /dt = ρ w 4/3 π R S 3. And finally, a radius increase rate is 2 ) NS π (R + R S ) 2 (v - v S ) E(R,R S ) using M = ρ w V = ρ w π 4/3 R 3 and dm /dt = ρ w 4 π R 2 dr /dt V. An integratable form of the continuous growth equation 1. we can start with the continuous growth equation a. The volume increase of the larger droplet is given as dv /dt = 4/3 π R S b. The mass increase of the larger droplet is then dm /dt = ρ w 4/3 π R S c. And finally, a radius increase rate is 2 ) NS π (R + R S ) 2 (v - v S ) E(R,R S ) using M = ρ w V = ρ w π 4/3 R 3 and dm /dt = ρ w 4 π R 2 dr /dt 2. Form 'c.' is most suitable 2 ) NS π (R + R S ) 2 (v - v S ) E(R,R S ) 5

6 3. Assumptions and algebra (nasty on a chalk board) a. R >> R S, (v >> v S ) b. From this we get (lets assume E = 1 continuous growth where E 1) dr /dt = 1/3 R S 3 N S π v c. multiply by 4/4 ρ w / ρ w ρ a/ ρ a (bear with me this is strange I know) d. From M = Vol ρ w =4/3 π R S 3 ρ w e. And we also know in general that mixing ratio of collected drops is q = M N / ρ air f. Now we have dr /dt = 1/3 R S 3 N S π 4/4 ρ w / ρ w ρ a/ ρ a g. using 'd.' above h. We get dr /dt = M R S 3 N S π v 1/4 1 / ρ w ρ a/ ρ a i. Using 'e.' from above dr /dt = 1/4 q v ρ a / ρ w 4. Now we can integrate this. et v = a R b R (t) R(0) 1 R b b dr dt = dt t t qa air 4 liq 1 1 b R 1 b (t) = 1 1 b R 1 b (t) + qa air t 4 liq dt R (t) = R 1 b (t) + (1 b) qa air t 4 liq a. a = 842 and b = 0.8 for rain 1/(1 b) VI. Various forms of the Collection equation There are three types of collection equations: 1) continuous, 2) quasi-stochastic, 3) pure-stochastic. Growth of droplets by these three equations is summarized in the pictorial in Fig. 17 from my notes (Fig. 7.7 in Young 1993). A. The continuous model (Young 1993) The continuous model predicts that all collector drops will have the same size after time interval dt. This implies that each droplet of radius r grows at the same rate. 1. Mathematically, the number dns (change in number of small droplets) is interpreted as the fractional number of small droplets collected by every collector of radius r in time interval dt. B. Quasi-stochastic model (Young 1993) The quasi-stochastic model predicts that all clouds will have the same size distribution after time interval dt. 1. Mathematically, the number dns is interpreted as the fraction of drops that collect a droplet in time interval dt. As collection is a discrete process, drops can not collect a fraction of a droplet, so we interpret this as saying a fraction f of drops will 6

7 collect droplets, and the remaining fraction (1-f) will not. 2. By this, drops initially the same size may have different growth histories, which allows for a spectrum of drops to form. However, only one outcome is permitted for each initial condition. 3. Two interpretations. a. Droplets maybe distributed uniformly i. A drop either collects a droplet, or does not collect a droplet. b. Droplets maybe distributed randomly i. Some droplets may collect one, two, or more droplets, while most others usually collect none. C. Pure-stochastic model (Young 1993) The quasi-stochastic model predicts that all clouds will have their own unique distributions after time interval dt. 1. Mathematically, the number dns is interpreted as the probability that a collector drop will collect a droplet in time interval dt. In this model it is assumed that droplets and drops have positions that are probabilistic in nature. D. The basic forms of the continuous growth equation: 1. The volume increase of the larger droplet is given as dv /dt = 4/3 π R S 2. The mass increase of the larger droplet is then dm /dt = ρ w 4/3 π R S 3. And finally, a radius increase rate is 2 ) NS π (R + R S ) 2 (v - v S ) E(R,R S ) using M = ρ w V = ρ w π 4/3 R 3 and dm /dt = ρ w 4 π R 2 dr /dt 4. Fig. 18 and 19 show results from integration of collisional growth equations. E. The basic form of the stochastic collection equation (SCE) k-1 dn k = dt 1 2 A i,k-i N i N k-i - N k A i,k N k i=1 i=1 where N is number concentration, and A is the collection kernel 1. The first sum is the gain term for size bin k a. Droplets of size k are produced by collisions of droplets in bin i, and k-i i. The factor of 1/2 accounts for the inherent double counting. For example, if k = 5, then droplets in bin k-i=3 (i=2) plus droplets in bin 2 make droplets of the size of those in bin k=5. Now, droplets in k-i=2 (i=3) plus droplets in bin 3 also make droplets of the same size as those in bin k=5 -- therefore, these interactions are double counted. 2. The second sum is loss term for size bin k a. Droplets of size k are lost by collisions of droplets in bin k with droplets in all other bins 3. The Fig. 20 attempts to make this clear. F. In general, stochastic growth equations are used to study the evolution of liquid droplet distributions and crystal aggregate distributions. 7 k-1

8 G. When the collector hydrometeor is much much larger than the collected hydrometeor, (hail collecting cloud droplets), the continuous growth equation is equally valid. VII. Collisional effects and drop break-up. These are summarized in Fig. 27. A. Neck or filament break-up. 1. Neck break up occurs by a glancing collision between a smaller and larger drop 2 As the smaller drop makes contact with the larger drop, a neck or filament forms. As the smaller drop continues. past the larger drop, the filament breaks. 3 The original drops retain much of their original mass. 4. Two to ten small fragments are formed (including original drops). The number of fragments increases with increasing collision energy. 5. Neck break-up occurs over a wide range of drop-size pairs. B. Sheet break-up 1. Sheet break-up occurs when the smaller drop collides with the larger drop such that it rips it into two pieces. 2. The onset of sheet break-up begins at larger small drop sizes than neck break-up. 3. The small drop is usually indistinguishable from the fragments. 4. arge drops is severely distorted from its original shape and is much smaller. 5. Two to ten small fragments are formed (including original drops). The number of fragments increases with increasing collision energy. C. Disk break-up 1. Disk break up occurs when the smaller strikes the larger drop along its center line. 2. After coalescence of the smaller and larger drop, a disk begins to form, extending to 2 or 3 times its original diameter. D. The aerodynamic forces then act to form a bowl shape particle which sheds droplets. The entire droplet may then break-up. 1. Few fragments are formed if the collisional energy is small. Up to 50 fragments may form if the collisional energy is high. 4. The number of fragments resulting from break up as a function of collisional energy is shown in Fig Fig. 29 shows typical size distributions produced after collisions of 2 and 3.5 mm sized droplets. E. Statistical breakup equation. 1. Based on the probability of break-up, a statistical break-up equation can be written... dn k dt = - N k P k + P k' Q k,k' N k' dm k' i=k 8

9 a. where i. m = mass ii. N(k) Number of drops with mass m(k) iii. P(k) probability of break-up of drop of mass m(k) per unit time P(k)=2.94x10-7 exp(34 r) iv. r = radius v. Q(k,k')dm(k') = number of drops with mass m(k) to m(k)+dm(k) formed by disintegration of one drop of mass m(k') Q(k,k')=145.37/m(k)[r(k) / r(k')] exp[-7r(k) / r(k')] (see Srivastava 1971, JAS) 9