Graupel and Hail Growth

Transcription

1 Graupel and Hail Growth I. Growth of large ice particles In this section we look at some basics of graupeln and hail growth. Important components of graupeln and hail growth models include production of embryos, diffusional growth, collection of ice and snow, collection of cloud and rain drops, and melting. The growth of larger ice particles is categorized as either dry growth or wet growth. A. Basics of dry growth The main requirement of dry growth is that all accreted mass is either frozen, or freezes completely after collection. 1. Vapor diffusion and accretion of liquid (which subsequently freezes) result in latent heat release. For dry growth, the temperature of the ice particle must remain below the freezing point of water so that the surface remains dry (solid ice). 2. The density of the added mass may be at a different density than that of the ice particle. For example, rime density can be as low as 170 kg/m 3 or as high as 917 kg/m 3 (solid ice). B. Basics of wet growth During wet growth, large ice particles collect significant amounts of supercooled liquid water, some of which does not freeze, because latent heat release warms the ice particle to the freezing point.. 1. Vapor diffusion and accretion of liquid (which subsequently freezes) result in latent heat release. For wet growth, the latent heat release that results from vapor diffusion and freezing of collected supercooled water is significant enough so that the ice particle surface temperature rises to K, the freezing point of water. 2. The unfrozen water can remain on the ice particle surface, soak porous ice (re-densification), or be shed as droplets. 3. The wet ice surface during wet growth results in efficient ice-ice sticking efficiency. This results in efficient mixed phase growth. 4. Wet growth is most likely for large ice particles in regions with larger liquid water contents (> 1 g/kg) and temperatures above -25 C. More will be said on this later. II. Heat budget equation for hydrometeors. To more accurately simulate the growth of large ice particles such as graupeln and hail a heat budget equation needs to be solved to determine particle temperature, and the amount of collected of supercooled liquid water that can freeze. A. The basic terms in the heat budget equation. The heat budget equation here is a sum of four terms... dq total /dt = dq 1 /dt + dq 2 /dt + dq 3 /dt + dq 4 /dt 1. Heat conduction term: dq 1 /dt dq 1 /dt = 2 π r f h K (T -T h ) r = radius f h is the ventilation coefficient for heating f h = X 2 x<1.4 f h = X x>1.4 1

2 x = N PR 1/3 N RE 1/2 T is the ambient temperature (K) T h is the hail surface temperature (K) 2. Heat of vaporization: dq 2 /dt dq 1 /dt = 2 π r f v L D v ρ a [q v, -q vs,h (T h )] L v is the latent heat of vaporization D v is the diffusivity of water vapor ρ a is the density of air f v is the ventilation coefficient for vapor f v = X 2 x<1.4 f v = X x>1.4 x = N SC 1/3 N RE 1/2 q v, is the ambient vapor mixing ratio q vs,h (T h ) is the saturation vapor mixing ratio at the hail surface temperature 3. Heat associated with accretion of liquid water: dq 3 /dt dq 3 /dt = dm l /dt [FF L f + C pw (T ) + C pi ( T h )] L f is the latent heat of freezing dm l /dt is the mass of liquid water accreted per unit time C pw is the specific heat of liquid water C pi is the specific heat of ice water FF is the fraction of accreted liquid that is frozen 4. Heat associated with accretion of ice water: dq 4 /dt dq 4 /dt = dm i /dt [C pi (T --T h )] dm i /dt is the mass of ice water accreted per unit time 5. The heat budget equation is then written as: dq total /dt = 2 π r f h K (T -T h ) + 2 π r f v L D v ρ a [q v, -q vs,h (T h )] + dm l /dt [FF L f + C pw (T ) + C pi ( T h )] + dm i /dt [C pi (T --T h )] B. Given r, dm l /dt, dm i /dt, T, q v,, we can find T h and FF from the heat budget equation by assuming a heat balance: dq total /dt = dq 1 /dt + dq 2 /dt + dq 3 /dt + dq 4 /dt = Solving this equation for T h involves applying an iterative technique like Newton - Raphson. The reason we need to do this is that q vs,h (T h ) is a function of T h. In solving this equation we assume that all of the liquid freezes (FF = 1.0). This gives us the maximum temperature the ice particle could have. 2. If T h > , we set T h = , and solve for FF, the fraction of frozen water. If T h 0, Then indeed FF = 1.0. Some researchers have noted that it might be a poor assumption to assume that the surface of an accreting an ice particle is 0 C. 3. The total mass accreted and frozen is then 2

3 dm wet-growth /dt = FF dm l /dt + dm i /dt 4. The amount of water that remains as liquid is: dm not-frozen /dt = (1-FF) dm l /dt, and can be absorbed into porous ice lattice, retained on the surface, or shed C. The Shumann-Ludlam limit describes the conditions for maximum rate at which ice particles can freeze all of the accreted supercooled liquid. A plot on the following page shows this limit. Ice particles growing beyond the Shumann-Ludlam limit grow by wet growth. These ice particles develop as spongy ice, in which unfrozen water is incorporated into the icewater matrix of the particle. If the ice fraction decreases below 0.7, drops begin to be shed. These particles are called spongy-shedding. Further decreases in the ice fraction, by enhanced wet growth, results in particles that shed all excess liquid not frozen so as to approximately maintain an ice fraction of 0.5. III. IV. Melting equation. The melting equation is in fact derived directly from the heat budget equation assuming the ice particle temperature is We usually assume that no ice is accreted. dm melt /dt = - (1/L f ) { 2 π r f h K (T -T h ) + 2 π r f v L D v ρ a [q v, -q vs,h (T h )] } - dm l /dt [C pw /L f (T -T h )] dm melt /dt is required to be negative. The ice temperature is assumed to be until it is completely melted. Modifications have been suggested for different Reynolds numbers regimes. A. When is melting of ice efficient? 1. Melting of ice is possible when air temperatures are warm (T>273.15) 2. Melting of ice is most efficient when the ambient is super-saturated with respect to liquid at K. This allows for condensation, and therefore, latent heating. Note that when the ambient T is above K, the wet ice surface temperature still can be K. The melting of ice can be slowed or essentially stopped if the wet bulb temperature (ambient) is below Melting of ice can be enhanced by accreting liquid water that is at the ambient temperature. Soaking of liquid water on hailstone surface during wet growth The soaking of a hailstone which becomes wet during WET GROWTH or MELTING is related to the porosity of the hailstone, and accessibility of porous regions of the hailstone from its surface. We can write a simple expression to state how much water might be absorbed assuming unlimited access to all porous regions Ṁmaxsoak = ρ w M h (1/ρ h - 1/ρ max ) where ρ max is typically 917 kg/m 3 if the soaked water freezes ρ w is the density of liquid water M h is the hail mass The hail density changes by accreting mass M soak of liquid (which can freeze) according to... ρ h ' = {ρ h M h + ρ soak M soak } / {M h + M soak } 3

4 V. Shedding water drops from hailstones. If a hailstone accretes more liquid than can freeze or soak, than it may remain on the surface or be shed. Experiments show that a liquid water torous forms on the hailstone surface, and if the hailstone is large than 9mm, it begins shedding droplets. A figure of this is shown on the attached page. The maximum amount of water that can remain on the hail surface is given by M surface-critical = M ice (mass in grams) If the mass of water on the hailstone surface exceeds M surface-critical it is shed. The modal size of the shed droplets is about 1.0 or 1.1 mm. VI. Density of hail A. Hailstone density is a function of the following 1. Embryo density a. frozen rain drop versus rimed crystal 2. Ambient and hailstone temperature 3. Density of mass collected - which is a function of ρ rime = 300 (-r w V imp / T h ) 0.44 kg/m 3 or ρ rime = 115 (-r w V imp / T h ) 0.76 kg/m 3 with a maximum of 917 kg/m 3 and a minimum of 170 kg/m 3 where the variables are as defined previously in Lecture 11. a. The density of regularly packed 30µm spheres each with a density of 1000kg/m 3 is 670kg/m 3 i. This is due to the air pockets between each of the frozen spheres. b. The density of other collected hydrometeors are (from PK81) i. crystals 570kg/m 3 ii. snow kg/m 3 iii. rain kg/m 3 iv. cloud drops kg/m 3 v. vapor?? (900kg/m 3 ) c. The density of a ice particle can after accreting other hydrometeors is mass weighted with the mass accreted Example... ρ h ' = {ρ h M h + ρ rime δt dm h /dt} / {M h + δt dm h /dt} d. Effect of hail density on hail terminal velocity.. i. Terminal velocity of a spherical hailstone Vt = { (4 g D ρ hail ) / (3 Cd ρ air ) } 1/2 ii. With diameter D = 2 cm, and ρ hail = 900kg/m 3 Vt = 28 m/s iii. With diameter D = 2 cm, and ρ hail = 500kg/m 3 Vt = 19 m/s e. Effect of hail density on hail growth 4

5 During hail growth, if a hailstones density can be reduced (e.g., by decreasing the riming density as a result of reducing the T hail, impact velocity, or droplet size), then the updraft requirements to keep the hailstone aloft are reduced as the terminal velocity of the hailstone is smaller. This might, in some instances, allow for a larger hailstone to grow. In addition, for the same mass, a lower density particle is larger than a higher density particle. However, production of lower density hailstones with smaller terminal velocities might not be able to remain in powerful updrafts for as long as higher density hailstones of the same mass. VII. Hailstone Embryos Hailstone embryos are typically ice particles less than 5 mm in diameter and have densities that can vary dramatically from 10 to 900 kg/m 3. Hydrometeors that can become hailstone embryos include the following. A. Graupeln Graupeln, which form by riming of small frozen drops, ice crystals, or snow flakes, can act as hailstone embryos. These embryos have densities that typically are about kg/m 3. B. Frozen droplets Frozen raindrops make the most dense hailstone embryos. Often they result from drops shed from melting hailstones. C. Spherical bubbly embryos. The origin of these embryos is unknown. They might be large rimed aggregates. They often are found in very large hailstones (D>5cm). D. Hailstone embryo climatology. Location % Graupel % Frozen drop S. Africa NHRE: D<25mm 88 8 NHRE: D>25mm Oklahoma: D<25mm Oklahoma: D>25mm Oklahoma: Switzerland: S. Africa: D<10mm S. Africa: 10<D<30mm S. Africa: 30<D<50mm S. Africa: D>50mm E. Hail size vs embryo type (this analysis is questionable) 1. In Colorado, most largest hail tends to form from frozen drops a. clouds have cold cloud base (T<15 C) 2. In Oklahoma, most largest hail tends to form from frozen drops a. clouds have cold cloud base (T<15 C) 3. In S. Africa, most largest hail tends to form from graupel particles VIII. Hailstone shape. Hailstones come in a large variety of shapes as depicted in the attached figures. A. Spherical hailstones Spherical hailstones often form from frozen drops. They accumulate loose rime, initially, causing them to tumble, and rime evenly. B. Conical hailstones 5

6 Conical hailstones form from riming of broke frozen drops or riming of crystals. C. Irregular shapes. Hailstones can form from aggregates of slushy graupel or snow. D. Lobes and protuberances. Lobes and protuberances can form on regions of hailstones with structural irregularities, which become favored growth sites. Protuberances are more common on very large hailstones. E. Hailstone aspect ratio. Hailstone aspect ratio tends to decrease slightly from 1.0 or 0.9 to 0.8 to 0.6 for increased hail sizes, and hailstones growth in wet growth regimes. F. Hailstones can fall vertically oriented, horizontally oriented, or with tumbling and gyrating motions. Vertically oriented hailstones might fall most stably. IX. Internal Structures in Hail. A. Air bubbles in hail Air bubbles in hail form as result of air being trapped in the water-ice lattice. The size of the bubbles depends, to some extent, on how fast the water freezes, and is related to whether wet or dry growth is occurring. The figures on the following page attempt to demonstrate the following findings. 1. Fast freezing. (dry growth?) Fast freezing of water results in many smaller trapped bubbles and results in 'cloudy' ice. Most bubbles are smaller than 10 µm. 2. Slow freezing. (wet growth?) Slow freezing results in fewer larger bubbles, and clearer ice. Most bubbles sizes are between 10 and 50µm in diameter. At warmer temperatures and during wet growth, there is a longer time between accretion and freezing, allowing smaller bubbles to aggregate. B. Ice Crystals Size in Hail. In general, fast freezing (cold temperatures) results in smaller crystal sizes, while slow freezing (warmer temperatures) results in larger crystal sizes. There is some correlation between crystal size and bubble concentration. Large crystals result in clear ice with low concentrations of large bubbles. Small crystals result in cloudy ice with high concentrations of small bubbles. The figures on the following pages attempt to compare theses. X. Conditions for large hail and large amounts of hail A. Large hail 1. Strong updraft 2. large amounts of supercooled liquid 3. light storm relative flow across large updraft 4. favorable horizontal updraft gradients 5. optimal embryo trajectories 6. large embryos 7. recycling 8. mixed phase growth B. Large amounts of hail (any size) 1. strong updrafts 2. large amounts of supercooled liquid 3. flow that injects many embryos across a broad updraft front 4. large number of embryos 6