The basic equation for the production of turbulent kinetic energy in clouds is. dz + g w

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Barrie Sims
5 years ago
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1 Turbuence in couds The basic equation for the production of turbuent kinetic energy in couds is de TKE dt = u 0 w 0 du v 0 w 0 dv + g w q 0 q 0 e The first two terms on the RHS are associated with shear production, the third with buoyancy production and the fourth with dissipation of TKE. Note the units, which are m 2 s 3, or W kg 1. Thus, dimensiona anaysis dictates that e u3 u2 t where u,, and t are characteristic speed, ength, and timescaes of the eddies, assuming that a revoution takes time t and u /t. The vaue of e can be estimate from the size of the argest eddy. For exampe for cumuus, it works out that e ranges from 10 4 to 10 2 W/kg, and for more stratiform couds e ranges from 10 6 to 10 4 W/kg. What happens to turbuent kinetic energy. Eventuay it is dissipated as heat due to viscous motions at very sma scaes. In between the argest scaes and smaest scaes, we think of an energy cascade where arge eddies break up into ever smaer eddies. Assuming that kinetic energy does not accumuate at any given scae, energy must enter one scae at the same rate it eaves it, hence e at arge scaes appies equay to a smaer scaes at east for the equiibrium condition. This does not mean however that the energy at a scaes is the same. Suppose we want to know the vaue of E(k) where k = 2p/ is the wavenumber. From dimensiona anaysis, we can proceed as foows: E (k) u 2 (e) 2/3 e 2/3 k Thus de dk e2/3 k 5/3 1

2 This -5/3 reationship is a fundamenta characteristic of isotropic turbuent fows. The cascade cannot continue forever, and eventuay it breaks down where viscous effects take over, i.e. where the Reynods number is given by Re u n 1 where n is the viscosity, 10 5 m 2 s 1 for air and is the characteristic Komogorov scae of these eddies. Since u (e 0 ) 1/3 it foows that e 1/3 4/3 0 n 1 or n 3 0 e 1/4 Note that is very weaky dependent on e so as a rue of thumb we can think of the smaest spatia scae of turbuence being about 1 mm. This is very interesting since it hints at the difficuty of modeing the dynamics of cumuus couds. Suppose that the argest eddies are 1 km, then this impies six orders of magnitude in ength, eighteen orders in voume! Further, the spacing between coud dropets is generay a bit arger than this scae, so it aso hints at the potentia for turbuence paying a roe in dropet coisions, a key component of rain production. Perhaps though we shoud confirm that a these eddies reay are turbuent. For this it is hepfu to refer to the Richardson number. Ri = du g q dq 2 + dv 2 The Richardson number expresses the static stabiity of the atmosphere reative to shear production of TKE that is avaiabe to overcome the static stabiity. If Ri is arge, say greater than 1 (though experimentay greater than about 0.25), then static stabiity is arge enough to suppress turbuence. 2

3 At the outer ength scae associated with the argest eddies with size and speed U, Ri b 0 U 2 2 b 0 U 2 1 where b 0 gq 0 /q and it was assumed that both shear and potentia energy compete equay at the outer ength scae. Hence Ri 1. At smaer ength scaes, we have Ri b 0 u 2 b 0! 1/3 2 U assuming e is constant, which reduces to Ri 1/3 so (perhaps not surprisingy), since < it foows that sma eddies are turbuent. What this tes us is that eddies of a sizes can be expected to be invoved with the entrainment of air between the cear air and its surroundings. We might suppose that the biggest eddies do most of the entraining, but is this reay true? There s aso a ot more sma eddies to consider. Perhaps we can take a stab at this question, again using dimensiona anaysis. What are the entrained voumes? This is easy V V 3 so big eddies are much bigger, and shoud gup much more. For exampe, suppose that 10, this is a factor of 1000 difference. But, the voume entrainment rate shoud have units of m 3 s 1 or E 2 U 3

4 so, since it foows that E E 1/3 u = U 2 U = u 7/3 This exponent is somewhat ess than 3, so for 10 we get a factor of 215 difference. But then there are aso many more sma eddies. et s suppose that sma eddies are on the surface of bigger eddies, in which case the number reevant to entrainment is determined by an area. n 2 Thus, we have for the entrainment ratio E ne 7/3 2 = 1/3 This is a very weak dependence! For the factor of 10 difference we mentioned in size, this woud transate ony to about a factor of 2 difference in entrainment rates. No particuar argument can be made that any particuar size contributes more. Entrainment matters at a scaes. What then is the timescae for mixing entrained air into the parce? et s assume the timescae for mixing is t mix u To determine this ratio we remember that de dk e2/3 k 5/3 or E u 2 e2/3 2/3 4

5 so t mix t mix U u 2/3 So this woud be a factor of 5 difference in timescaes for mixing per factor of 10 difference in spatia scae. et s ook at these timescaes. If t mix U at the argest scaes, then say for cumuonimbus is 1000 m and U is 10 m/s. Then t mix 100 s. This shoud not be that surprising since it is roughy the inverse of N the Brunt-Vaisaa frequency. At the smaest scaes of 1 mm, it foows that the mixing time scae is reay fast, about 0.01 s. So we re ooking at about four orders of magnitude in mixing time scaes. 5

THINKING IN PYRAMIDS

ECS 178 Course Notes THINKING IN PYRAMIDS Kenneth I. Joy Institute for Data Anaysis and Visuaization Department of Computer Science University of Caifornia, Davis Overview It is frequenty usefu to think