Presentation is loading. Please wait.

Presentation is loading. Please wait.

Assumes supersaturation=0.05%, p=900 hPa, T=273K r dry = 0.1 0.22 0.48  m.

Similar presentations


Presentation on theme: "Assumes supersaturation=0.05%, p=900 hPa, T=273K r dry = 0.1 0.22 0.48  m."— Presentation transcript:

1

2

3 Assumes supersaturation=0.05%, p=900 hPa, T=273K r dry = 0.1 0.22 0.48  m

4

5 Tropical profile Polar profile THERMAL DIFFUSION LIMITED VAPOR DIFFUSION LIMITED

6

7

8

9

10

11 ICE FORMATION AND CONDENSATIONAL GROWTH

12

13

14

15

16 PRECIPITATION

17 Terminal velocity of drops Balance between gravitational force (= mg) and drag force (  size) Time taken for drop to reach terminal velocity is very short radius [µm] terminal velocity [m s -1 ] r2r2 rr  r 1/2 “non precipitating” “precipitating”

18

19

20

21

22 All collisions Collisions only for S 1 drops: “autoconversion” Collisions between S 1 and S 2 drops only: “accretion” Collisions only for S 2 drops only: “self-collection”

23 Ice crystal aggregates in Cirrus clouds Taken using an imaging probe on an aircraft Most crystals show irregular aggregated shape Snow in midlatitudes mostly grows by aggregation Westbrook et al. 2004

24

25


Download ppt "Assumes supersaturation=0.05%, p=900 hPa, T=273K r dry = 0.1 0.22 0.48  m."

Similar presentations


Ads by Google