1/2 and e0 e s ' 1+ imm w 4 M s 3 πρ0 r 3 m. n 0 ktr. .Also,since n 0 ktr 1,wehave. 4 3 M sπρ 0 r 3. ktr. 3 M sπρ 0

Transcription

1 Chapter Shw that fr a very weak slutin drplet (m 4 3 πr3 ρ 0 M s ), (6.8) can be written as e 0 ' 1+ a r b r 3 where a σ 0 /n 0 kt and b imm w / 4 3 M sπρ 0. What is yur interpretatin f thecnd and third terms n the right-hand side f this expressin? Shw that in this case the peak in the Köhler curve ccurs at Slutin r ' Frm eqn. (6.8) in text 3b a 1/ and e0 ' 1+ e 0 exp σ0 n 0 ktr 1+ 4a 3 7b 1/ imm w 4 M s 3 πρ0 r 3 m Fr a very weak slutin m 4 3 πr3 ρ 0 σ 0.Als,since n 0 ktr 1,wehave r, e 0 l 1+ σ0 n 0 1 ktr l 1+ σ0 n 0 ktr e 0 imm w 4 3 M sπρ 0 r 3 imm w [ ] 4 3 M sπρ 0 r 3 1 r 4 1 {z } very small l 1+ a r b r 3 (1) where, a σ0 n 0 kt and b imm w 4 3 M sπρ 0 Thecnd term n the right side f (1) represents the effect f the curvature f the drp in increasing e 0, and the third term represents the effect f disslved salt in decreasing e 0 Á. The Köhler curve (i.e., e 0 Á versus r) reaches a peak value when d dr e 0 0, 3

2 that is, when r, when 0 l a r 3b r 4 r l 3b a The maximum value f e0 is e 0 ³ a 1/ ³ a l 1+a b max 3b 3b 3/ a 3 1/ b /3 a l 1+ 3b 3b a 3 1/ a 3 1/ l 1+ 3b 3 3 b a 3 1/ l 1+ 1 a 3 1/ 3b 3 3b l 1+ a 3 1/ 3 3b e 0 max l 1+ 4a 3 7b 1/ 3/ 6.15 The air at the 500 hpa level in a cumulnimbus clud has a temperature f 0 C and a liquid water cntent f 3 g m 3. (a) Assuming the clud drps are falling at their terminal fall speeds, calculate the dwnward frictinal drag that the clud drps exert n a unit mass f air. (b) Express this dwnward frce in terms f a (negative) virtual temperature crrectin. (In ther wrds, find the decrease in temperature the air wuld have t underg if it cntained n liquid water in rder t be as dense as the air in questin.) [Hint: At temperatures arund 0 C the virtual temperature crrectin due t the presence f water vapr in the air can be neglected.] Answer (a) N kg 1. (b) 1.8 C. 33

3 Slutin Figure (a) When drps are falling at terminal fall speed, the frictinal drag per unit mass f air is F d Dwnward frce n drps/unit mass f air mass f drps g mass f air Since, air density ρ Als, fr air F d mass f air vlume f air (mass f drps) (vlume f air) g ρ kg m 3 g ρ p R d ρt F d g R dt p (9.81) (87) (73) F d N kg 1 (b) Withut drps being present Dwnward frce acting n a unit mass f air g 34

4 When drps are present Dwnward frce acting n a unit mass f air (g ) ( ) N kg 1 But, Figure Pressure at grund p g p g Z dp dz ρg ρgdz g Z Z p g dp ρdz (1) Nw the presence f 3 g m kg m 3 f water in air, will have negligible effect n the density f air, which is 1.75 kg m 3 at surface. Therefre frm (1) we can write p g g Similarly, at clud height Hence, p g p (air alne)) p (air + clud water) g g Since, p R d ρt the air in questin (i.e., air cntaining water vapr) has density ρ p (air + clud water) R d T () (3) 35

5 If the air had n clud water, but is required t have density ρ given by (3), then p (air nly) Density R d T v r, p (air + clud water) p (air alne) R d T R d T v where T v is the virtual temperature required if the air alne is t have density ρ. Frm () and (4) T v p (air alne) p (air + clud water) T (4) T v T (73) 71.7 K the (negative) virtual temeprature crrectin is C. 6.0 Cnsider a saturated parcel f air that is lifted adiabatically. The rate f change in thupersaturatin rati S (S e/ where e is the vapr pressure f the air and thaturated vapr pressure) f the air parcel may be written as Q dz 1 Q d(lw C) where dz/ and LW C are the vertical velcity and the liquid water cntent f the air parcel, respectively. a) Assuming that S is clse t unity, and neglecting the difference between the actual temperature and the virtual temperature f the air, shw that Q 1 ' g εlv 1 TR d Tc p where, T is the temperature f the parcel (in degrees kelvin), L v the latent heat f cndensatin, g the acceleratin due t gravity, R d and R v are the gas cnstants fr 1 kg f dry air and 1 kg f water vapr, respectively, ε R d /R v,andc p thpecific heat at cnstant pressure f thaturated air. [Hint: Cnsider ascent with n cndensatin. Intrduce the mixing rati f the air, which is related t e by Eqn. (3.6).] b) Derive a crrespnding apprximate expressin fr Q in terms f R d, T, e,, L v, p, c p and the density f the mist air (ρ). Assume that e/ ' 1, T ' T v and ε>>w(the mixing rati f the air). 36

6 Rd T Answer Q ρ ε Slutin + L v pt c p (a) Fr ascent with n cndensatin LWC 0, therefre: since, Q dz 1 (1) We will first evaluate de S e Á de ed e s () e w ε + w p where, w mixing rati, which is cnstant if there is n cndensatin. r But, and Frm (3) (4): de w dp ε + w de w ε + w dp dz gρ dp dz dz (3) p R d ρt (T T v ) (4) de eg dz R d T We will nw evaluate d. Frm the Clausius-Clapeyrn equatin d dt L v T (α α 1 ) ' L v T (α ) (5) and, R v 1 α T d dt L v R v T 37

7 r, But, d L v dt R v T d L v dt dz R v T dz dt dz +g c p Dry adiabatic lapse rate (because there is n cndensatin) Frm (), (5) and (6): d L v g dz R v T c p (6) 1 T e εlv g R d c p T g R d dz Since e ' 1 ' 1 εlv g T R d c p T g dz R d Cmparing (1) and (7): Q 1 g εlv TR d c p T 1 dz (b) If we assume n vertical air mtin cndensatin ccurs, then: Q d (LW C) Equatin () still hlds, and we will nw evaluate de e w ε + w p w nw varies, but p is cnstant. But, de ε dw (ε + w) p (7) ' 0 and p cnstant while (8) fr this case. 38

8 dw (LW C) d de ε (LW C) pd (ε + w) Since p R d ρt (T T v ) and ε À w de ' 1 ε ρr d (LW C) dt (9) We nw evaluate d fr this case. As befre, frm the Clausius Clapeyrn eqn. and the gas eqn. fr vater vapr, we get: But, and als, Hence, r, d Frm (10) and (11): ' L v dt R v T L v R v T dt d (LW C) d (LW C) dw d (LW C) dq L v d (LW C) L v dw dq c p dt c p dt L v dw (10) dt L vdw c p (11) d + L v d (LW C) R v T c p (1) Frm (), (9) and (1): Á de ed e s 1 d (LW C) ρr d T e L v d (LW C) ε R v T c p 39

9 Substitute p ρr d T s that T p/ R d ρ Or, since e ' 1, 1 d (LW C) ρr d T e L vr d ρ d (LW C) ε R v Tc p p ' ρ Rd T ε + εl v d (LW C) Tpc p (13) Frm (8) and (13), Rd T Q ρ ε + εl v pt c p 6.4 If a raindrp has a radius f 1 mm at clud base, which is lcated 5 km abve the grund, what will be its radius at the grund and hw lng will it take t reach the grund if the relative humidity between clud base and grund is cnstant at 60%? [Hint: Use (6.1) and the relatinship between v and r given in Exercise 6.3. If r is in micrmeters, the value f G l in (6.1) is 100 fr clud drplets, but fr the large drp sizes cnsidered in this prblem the value f G l shuld be taken as 700 t allw fr ventilatin effects. Answer 0.67 mm and 16.4 min Slutin Frm the relatin between v and r given in Exercise (6.3) V r where V is in m s 1 and r in meters. V r 0.6r where V is nw in cm s 1 and r is in m. Let x distance f drp (in cm) belw clud base at time t, then dx where r is in m. V 0.6r (1) dx 0.6r Frm (1), r dr G S 7 10 S () 40

10 where r is in m an is expressed as a fractin. rdr 7 10 S dx 0.6r r, 7 10 r S dr dx 0.6 integrating frm clud base, where r 1000 m andx 0,t the grund, where r R (in m) and x cm, Z R(m) 1000 m r 3 3 r dr 700S 0.6 R(m) S 0.6 Z cm 0 cm If RH 60% thupersaturatin (as a fractin) is e e 1 RH (as a fractin) dx S 0.4 r 3 3 R(m) 1000 R ( 0.4) 0.6 7(0.4) (5) R Time taken, frm () abve: R m Radius at clud base 0.67 mm rdr 700S 41

11 r, But, frm (1) abve, Hence, Z r 1000 m r 1000 r 1000 rdr 700S r Z t 700St 700St r (700) St dx 0.6r dx St 1/ Z cm Z T dx St 1/ St T 3/ S ST 3/ S 400S But S 0.4, therefre, T 3/ ( 0.4) 400 ( 0.4) ( 0.4) 9 / T 1. r, T secs T16.4 mins 6.5 A large number f drps, each f vlume V, are cled simultaneusly at a steady rate β( dt/). Letp(V,t) be the prbability f ice nucleatin taking place in a vlume V f water during a time interval t. (a) Derive a relatinship between p(v,t) and R T t J 0 LS dt,wherej LS is the ice nucleatin rate (per unit vlume per unit time) and T t the temperature f 4

12 the drps at time t. (b) Shw that a n-fld increase in the cling rate prduces thame depressin in the freezing temperature as an n-fld decrease in drp vlume. Answer ln[1 p(v,t)] V β Slutin R Tt 0 J LS dt (a) Let N ttal number f drps N t number f drps frzen at time t Then P (V,t) N t N Number f drps that nucleat between time and t + is Dividing bth sides by N N t+ N t +(N N t ) VJ LS P (V,t + ) P (V,t)+[1 P (V,t) VJ LS ] Since P (V,t + ) P (V,t)+ d [P (V,t)] it fllws that Hence, d P (V,t)[1 P (V,t)] VJ LS Z P (V,t) But β dt, therefre dp (V,t) 1 P (V,t) ln [1 P (V,t)] Z t Z t VJ LS VJ LS ln [1 P (V, t)] V β where T t is temperature at time t. Z Tt J LS dt 43

13 (b) Frm the equatin fllwing (6.34) in Exercise (6.4): n ln(1 P ) ln (1 P ) 4 3 D 3 π 10 3 exp a (T 1 T ) Since frm (a) abve ln (1 P ) V β V β exp a (T 1 T ) V and β have inverse effect n the medium freezing temperature T If the velcities f sund in tw adjacent thin layers f air are v 1,and v, a sund wave will be refracted at the interface between the layers and sin i sin r v 1 (see Fig. 6.65). v Figure Use this relatinship, and the fact that v (T ) 1/,whereT is the air temperature (in degrees kelvin), t shw that the equatin fr the path f a sund wave prduced by thunder that is heard at the maximum hrizntal distance frm the rigin f the thunder is dx (T/Γz) 1/ dz 44

14 where x and z are the crdinates indicated in Fig. 6.66, Γ the temperature lapse rate in the vertical (assumed cnstant), and T the temperature at height z. Slutin Or, in general, But, Frm Fig and the law f refractin sin φ v sin i sin r v 1 v (1) cnstant (alng any path f a sund wave) () tan Φ dx dz dx dz sin Φ cs Φ sin Φ 1 sin Φ dx 1/ (3) At pint P in Fig. E6.37 Figure E6.37. sin 90 v cnstant 45

15 r, 1 cnstant (4) v where, v is velcity f sund just abve grund level. Hence, in general, frm () sin Φ 1 v v But v T, therefre, Frm (3) and (5) sin Φ v 1/ T (5) v T dx dz (T / T ) 1/ 1/ T 1/ 1 TT T T But, and, T T Γz T T Γz T 1/ dx dz Γz 6.38 Use the expressin derived in Exercise (6.37) t shw that the maximum distance D at which a sund wave prduced by thunder can be heard is given apprximately by (see Fig frm Exercise 6.37) D (T 0 H/Γ) 1/ where T 0 is the temperature at the grund. Calculate the value f D given that Γ 7.50 Ckm 1, T K, and H 4km. Answer 5.3 km 46

16 Slutin Frm Exercise 6.67: 1/ T Γz dx dz Γz T dx ' 1/ Γz 1 dz 1/ 1 Γz T Γz T Γz T Γz T 1/ 1 1 Γz 1/ 1 1 1/ dz T + Γz T dz higher pwered terms in z dz Integrating, Z D dx r Z T T Γ H 1/ Z Γ T Γ T Γ T Γ T H D Γ 1 z 1 H Γz dz T 1 z 1 1/ z 1/ 1/ z 1/ Γz dz T Γ T 3 z3/ Γ 3T z 3/ H 1/ H 1/ + 1 Γ H 3/ 3 T {z } small term 1/ With Γ 7.5 km 1, T 300 K, and H 4km D D 5.3 km 1/ H 47