Atsc. 405 2012 final Equations: page 1/6 Answer each of these 7 questions (note weight). Show all your work on all questions (needed for partial credit). Be sure to put your name on any detached pages. 1. (10) Starting from (33) show that for a flat sheet of liquid water in equilibrium with vapor, g l = g v. According to (33): dg sdt + αdp (1) where we are talking about the mixture of liquid and vapor and specifying that water is neither entering or leaving the system. If dt=0 and dp=0 then dg can only be negative or zero, so in equilibrium g=constant. But we also know that :G = m v g v + m l g l and since neither g v = h v T s v and g l = h l T s l will change if T and p are constant, we ve got: dg = 0 = g v dm v + g l dm l (2) Next, since water is conserved we know that dm v = dm l, plug those in and we get g l = g v 2. (12) Suppose you measure a drizzle size distribution with radii between 0.2-2 mm that takes the form: n(r) = 700r 3 (3) where n(r) is the number density in m 3 mm 1. For the 0.2-2 mm size range find The total number concentration N in m 3 In [4]: -0.5*(700)*((1./r[1])**2. - (1./r[0])**2.) Out[4]: 8662.5 #m^-3 The mean radius r in mm In [6]: (-700)*((1./r[1]) - (1./r[0]))/8662.5 Out[6]: 0.36365735395982451 The rain rate, in mm/hr, given a fall speed of V (r) = 6r, where V (r) is in m/s and r is in mm. The rainrate R in meters/second is:
Atsc. 405 2012 final Equations: page 2/6 R = 4 2 3 π r 3 10 9 700r 3 6rdr (4) 0.2 = 4 3 109 4200.π r2 2 2 0.2 = 4 3 109 4200. 0.5π (4 0.04) m/s = 3.4834 10 5 m/s 3600 1000. = 125.4 mm/hr (5) Matlab script to do this numerically spacing=0.0001; %mm r=0.2:spacing:2;%mm number=700*(r).^(-3).*spacing;% m^{-3} concentration in each bin tot_num=sum(number); out_mesg={ \ntotal number is %8.4f m^{-3}\n }; fprintf(strcat(out_mesg{:}),tot_num); mean_rad=sum(number.*r)/tot_num; out_mesg={ \nmean radius is %8.4f mm\n }; fprintf(strcat(out_mesg{:}),mean_rad); Vfall=6*r; %m/s rmeter=r*1.e-3; % meters Volume=4./3.*pi*rmeter.^3.; R=sum(number.*Volume.*Vfall); %m/s R=R*1000*3600.; %mm/hour out_mesg={ \nrain rate is %8.2f mm/hour\n }; fprintf(strcat(out_mesg{:}),r); When run this prints out: total number is 8666.8805 m^{-3} mean radius is 0.3636 mm rain rate is 125.41 mm/hour 3. (6) Use Taylor series to show that: log θ e 1 θ e (6) θ e for sufficiently small θ e. let y = log θ e so that y (θ e ) = 1 θ e now expand about θ e1 using a first order Taylor series:
Atsc. 405 2012 final Equations: page 3/6 y 2 = log θ e2 = log θ e1 + 1 θ e1 (θ e2 θ e1 ) (7) log θ e2 log θ e1 = log θ e = 1 θ e θ e (8) 4. (15) Moist adiabat Describe how you would calculate a set of (temperature, pressure) coordinates that could be used to draw a single moist adiabat on a tephigram using matlab. Do it in two ways: ( ) (a) With fzero: use the relationship θ e = θ exp l vw s c pt (first show how integrating equation (25) gives this expression). (b) With ode45: Derive an equation for dt/dp from (24), assuming hydrostatic balance and also assuming that you can call functions that will give you w w and dw s /dt for any temperature and pressure. In each case describe how you would start and stop the calculation, and write out the routines in pseudo-code that doesn t have to run, but should translate roughly line for line to Matlab. 5. (6) Conservation laws Prove using (19) that the moist static energy h m defined by (24) is conserved for adiabatic motion in a hydrostatic atmosphere. Use this to show that h m for a mixture of two parcels is just the sum of the moist static energies of the two parcels: h mc = h ma + h mb.
Atsc. 405 2012 final Equations: page 4/6 6. (11) Köhler curve The figure shows two droplets with radii of 0.2 µm and 1 µm at a environmental relative humidity of about 1.001. (a) (3) Both droplets lie on a curved line called the Koehler curve. What do points on this line represent? i.e. what is it that all points lying on this line have in common? (b) (4) Suppose the relative humidity increases from 1.001 to 1.002. Describe qualitatively what happens to the two drops. (c) (4) Quantitatively, what is the initial evaporation/growth rate of droplet 2 in microns/second, assuming a RH of 1.002, temperature T=280, esat(t)=10 hpa. 7. (6) Show with the help of a figure why: ρ v t = D 2 ρ v (9)
Atsc. 405 2012 final Equations: page 5/6 Equation sheet du = q dt w dt = q dt p dα (10) dw v = w ( v p ) p e e de dp (27) e = ρ v R v T (11) p = ρ R d T v (12) f(x) = f(x 0 ) + f (x 0 )(x x 0 ) + f (x 0 ) (x x 0 ) 2 +... (28) 2 w dt = p dα (13) h = u + p α (14) T v = T (1 + 0.608w v w l ) (15) l v = T (s v s l ) (29) de s dt = l ve s R v T 2 (30) e s w vs = ρ s /ρ d = ɛ (16) p e s dh = c px dt (dry air or liquid) (17) dh = c p dt + l v dw v (air/water mixture) (18) h l = c l (T T p ) h v = l v0 + c pv (T T p ) s d = c pd log T R d log p d s l = c l ln T T p s v = c pv ln T R v ln e + l v0 T p e s0 T p (31a) (31b) (31c) (31d) (31e) dh = T ds + α dp (reversible) (19) dθ ds = c p θ = c dt p T R dp d p ds q dt T (20) (21) l v = h v h l (22) e g = u + pα T s = h T s (32) dg sdt + αdp (33) E = m v g v + m l g l + 4πσr 2 (34) ( ) S d(g l g v ) R v T ln a we s a w (35) dp = ρ g dz (23) dh m = c p dt + l v dw v + g dz (24) e r = e s (T )(n w /(n w + n s )) exp(a/r) = e s (T )(1 + a r b r 3 ) (36) ds = c p dθ es θ es ds = c p dθ l θ l dθ = c p θ + l v dw s T dθ = c p θ l v dw l T (25) (26) ( ) 4a 3 1/2 s crit = 1 + 27b (37) ( ) 1/2 3b r crit = a (38)
Atsc. 405 2012 final Equations: page 6/6 c p dθ e θ e = dm m h m T = dm m s = c p dm m log θ e (39) 1 dm m dt = λ (40) dw T dz = (w w T )ˆλ (41) dm = πr 2 E c (V (R) V (r))w l (42) dt where V(r) 0 and V(R) JR (with J=6000 s 1, Rin m) dm dt = 4πrD(ρ v ρ vr ) (43) dr dt = 1 Dρ v [e e r ] (44) r ρ l e n dm dt = DC ɛ 0 (ρ v ρ vc ) (45) N(D ) = D n(d)dd (46) 2σ a ρ l R vt imm b w (4/3)M sπρ σ 0.075 J m 2 ρ l 1000 kg m 3 c pd 1006 J kg 1 K 1 c vd 719 J kg 1 K 1 c pv 1870 J kg 1 K 1 c vv 1408 J kg 1 K 1 c l 4190 J kg 1 K 1 D 2.36 10 5 m 2 s 1 R d 287 J kg 1 K 1 R v 461 J kg 1 K 1 k 1.381 10 23 J K 1 molecule 1 l v0 2.501 10 6 J kg at 0 deg C