Atmosphere, Ocean and Climate Dynamics Answers to Chapter 4

Transcription

1 Atmoshere, Ocean and Climate Dynamics Answers to Chater 4 1. Show that the buoyancy frequency, Eq.(4.22), may be written in terms of the environmental temerature rofile thus N 2 = g µ dte T E dz + Γ d where Γ d is the dry adiabatic lase rate. Using the definition (4.17) for the environmental otential temerature θ E,Eq.(4.22)gives N 2 = g dθ E θ E dz = g T E = g T E µ E " κ dt E µ dz E dte dz κt E d E E dz κ κ µ. E κ 1 T E d E dz But, from the ideal gas law (1.1), T E / E =(Rρ E ) 1, while hydrostatic balance (3.3) we have d E /dz = gρ E, whence N 2 = g dte T E dz + gκ = g dte R T E dz + Γ d, where Γ d = gκ/r = g/c is the adiabatic lase rate. 2. From the temerature ( T )rofileshowninfig.4.9: (a) estimate the troosheric lase rate and comare to the dry adiabatic lase rate. The mean temerature gradient, with resect to ressure, over the layer 7 3h Pa is aroximately T/ ' 45 K/4 h Pa ' KPa 1. Using the ideal gas law (1.1), a mean temerature (see art (b) of this question) of 26 K and a mean ressure =5h Pa, the mean density of the layer is ρ o = 1 #

2 o / (RT o )=5 1 4 / (387 26) =.5 kg m 3. Hence, using hydrostatic balance (3,3), T z = T z = gρ T ' ' Km 1, i.e., the mean lase rate is about 5.4 Kkm 1,comaredwiththe adiabatic lase rate Γ d = g/c =9. 8Km 1. So this temerature rofile is stable to unsaturated convection. (b) estimate the ressure scale height RT /g, wheret o is the mean temerature over the 7 mbar to 3 mbar layer. Estimating the mean temerature in the layer to be T o ' 26 K, the corresonding scale height is H = RT o /g =287 26/9.81 = 7.6 km. (c) estimate the eriod of buoyancy oscillations in mid-trooshere. Using the result from Q.1, the buoyancy frequency N is given by N 2 = g dte T E dz + Γ d. Substituting from our estimates, N 2 ' [ ] 1 3 = s 2, whence the buoyancy eriod is 2π/N ' 2π/ ' 48 s. 3. Consider the laboratory convection exeriment described in section The thermodynamic equation (horizontally averaged over the tank) can be written: dt ρc dt = H (4.31) h where h is the deth of the convection layer see Fig.4.28 H is the (constant) heat flux coming in at the bottom from the heating ad, ρ is the density, c is the secific heat,t is time and T is temerature. We observe that the temerature in the convection layer is almost homogeneous and joins on to the linear stratification into which the convection is burrowing, as sketched in Fig Show that if this is the case, Eq.(4.31) can be written thus: ρc T z 2 2 d dt h2 = H

3 where T z isassumedtobeconstant. If the stratification is linear (T z = constant) and the temerature of the boundary layer joins on to it (as sketched in the figure) then the temerature of the boundary layer is directly related to its deth h and, to within a constant, can be written, T = T z h.thus,eq.(1)can be rearranged to yield the equation above. Solve the above equation for h(t) and hence show that the deth and temerature of the convecting layer increases like t. Sketch the form the solution for h(t) and T (t). Solving we find that: s h 2 = 2H 2H t or h = t ρc T z ρc T z if H and T z are constant. Thus h and, in view of T = T z h, T both increase like t,assketchedhere. h or T (normalized) time (a) Is your solution consistent with the lot of the temerature evolution from the laboratory exeriment shown in Fig.4.8? Yes, the exerimental data is consistent with a t deendence. (b) How would T have varied with time if initially the water in the tank had been of uniform temerature (i.e. was unstratified)? You may assume that the water remains well mixed at all times and so is of uniform temerature. 3

4 In this case the deth of the fluid undergoing convective mixing is constant and Eq.(4.31) therefore imlies that T increases linearly with time. 4. Consider an atmosheric temerature rofile at dawn with a temerature discontinuity (inversion) at 1km, andatrooause at11km, such that 1 o C, z<1km T (z) = [15 8(z 1)] o C 1 <z<11km 65 o C z>11km (where here z is exressed in km). Following sunrise at 6 a.m. until 1.m., the surface temerature steadily increases from its initial value of 1 o Catarateof 3 o C er hour. Assuming that convection enetrates to the level at which air arcels originating at the surface attain neutral buoyancy, describe the evolution during this time of convection (a) if the surface air is comletely dry; (b) if the surface air is saturated. [You may assume that unsaturated (saturated) air arcels follow the dry (moist) adiabatic lase rate of 1 K km 1 ( 7K km 1 ) under all conditions (even at finite dislacements).] a) As the ground starts to warm after dawn (temerature T s =(1+3t) C, where t is time in hours after 6am), arcels from the surface will be ositively buoyant until they reach an altitude z to (km) such that T s 1z to = T init. z to will reach 1km when T s =2 C, at 9:2am. Until that time, convection cannot extend above a level z to, which is less than 1km. (E.g., at 8:am, T s =16 C, z to =.6km.) z to remains at 1km until T s =25 C, at 11:am. After that time, convection will extend beyond 1km, in fact to z to,where T s 1z to =23 8z to, i.e., z to = 1 2 (T s 23). 4

5 Thus, the deth of convection will gradually increase as T s increases (since the background lase rate above 1km, 8K km 1, is greater than Γ d ), reaching 4km at 1:m, when T s =31 C. b) The maximum height of convection, z to (the level of neutral buoyancy), is now defined by T s 7z to = T init. Qualitatively, the early develoment of convection is similar to that of the dry case. Convection first reaches 1km when T s = 17 C, at 8:2am, and remains at that level until T s =22 C, at 1:am. After 1:, however, the situation changes raidly. If T s > 22 C then, after the arcel asses 1km altitude, it remains ositively buoyant all the way to the trooause, because the environmental lase rate above 1km is conditionally unstable (dt/dz < Γ s ). In fact, after 1:, the level of neutral buoyancy is given by T s 7z to = 65. Just after 1: (T s =22 C), therefore, convection extends all the way to 12 3 km. The reason for the dramatic difference from the 7 unsaturated case is that the lase rate above the inversion exceeds the saturated lase rate: the rofile is conditionally unstable. Thus, once surface air becomes warm enough to enetrate the inversion, it rises all the way to the trooause. After this time, z to increases slowly to a value 13 5km at 1:m, when T 7 s =31 C. 5. For a erfect gas undergoing changes dt in temerature and dv in secific volume, the change ds in secific entroy, s, is given by Tds= c v dt + dv. (a) Hence, for unsaturated air, show that otential temerature θ µ κ θ = T, (notation defined in notes) is a measure of secific entroy; secifically, that s = c ln θ + constant. 5

6 where c v and c are secific heats at constant volume and constant ressure, resectively. Change in entroy, ds, given by Tds= c v dt + dv where secific volume change dv = d(1/ρ) =d(rt/) = RdT 1 d. So ρ dv = RdT 1 ρ d. Hence Tds= c dt 1 ρ d, where c = c v + R, andso Now, otential temerature is ds = c dt T Rd. (1) θ = T µ κ, where is a constant. Therefore µ κ dθ = dt κt µ κ d or dθ θ = d ln θ = dt T κd. So dt c d ln θ = c T Rd, (2) since κ = R/c. Comaring (1) and (2), or s = c ln θ+constant. ds = c d ln θ, 6

7 (b) Show that if the environmental lase rate is dry adiabatic, then it has constant otential temerature. Fromart(a)wehaves = c ln θ+ constant, thus ds = c d(ln θ) = dθ c. When the arcel is moved adiabatically, no heat is exchanged θ with the surroundings - dq = Tds =- its entroy is conserved. dθ From the relation ds = c d(ln θ) =c,we see that dθ =, i.e. θ the otential temerature is conserved when the arcel is moved adiabatically. ³ 6. Investigate under what conditions we may aroximate L dq c T by d Lq c T in the derivation of Eq.(4.3). Is this a good aroximation in tyical atmosheric conditions? The key assumtion made in the derivation of (4.3) is that, in the exression µ Lq d = L Lq dq c T c T c T dt, 2 the second term may be ignored comared to the first. If we rewrite the equation as µ Lq d = Lq c T c T µ dq q dt T we can see that the neglect of the second term is tantamount to assuming that δ (ln T ) δ (ln q). Now, from the global mean moisture rofile shown in Fig. 3.3, over the altitude range from the surface to 4 hpa q decreases from about 17 to about 1g kg 1,whence δ (ln q) 'ln (.17) ln (.1) ' 2.8, whereas, over the same height range, T varies from about 288 to about 253 K, whence δ (ln T ) ' ln (288) ln (253) '.13. Hence, for this examle, neglect of the second term results in a fractional error of a little less than 5%. 7. Assume the atmoshere is isothermal with temerature 28K. Determine the otential temerature at altitudes of 5km, 1 km, and2 km above the surface. If an air arcel were moved adiabatically from 1km to 5km, what would its temerature be on arrival? For an isothermal atmoshere with temerature T, ressure varies with altitude as ³ (z) = s ex z, H 7,

8 where s is surface ressure (1hPa) and the scale height is H = RT/g (in our usual notation). Given T =28K, H = Potential temerature is m =8.192km θ = T µ κ s, where κ =2/7. z = 5km : (5km) = 1 ex( )=543hPa. Therefore θ(5km) = = 333K. 543 z = 1km: (1km) = 1 ex( )=295hPa. θ(1km) = = 397K. 295 z = 2km: (2km) = 1 ex( )=87hPa. µ θ(2km) =28 =563K. 87 In an air arcel is moved adiabatically from 1km to 5km, it will conserve its otential temerature. Therefore, on arrival at 5km, it will have the same θ as when it left 1km, i.e., 397K. Its temerature on arrival is therefore µ κ µ T = θ =397 =333K =6 C(!). s 1 8

9 8. Somewhere (in a galaxy far, far away) there is a lanet whose atmoshere is just like that of the Earth in all resects but one it contains no moisture. The lanet s trooshere is maintained by convection to be neutrally stable to vertical dislacements. Its stratoshere is in radiative equilibrium, at a uniform temerature 8 C, and temerature is continuous across the trooause. If the surface ressure is 1hPa, and equatorial surface temerature is 32 C, what is the ressure at the equatorial trooause? If the trooshere is neutrally stable to dry convection, then otential temerature θ is uniform with height within the trooshere. Since θ = T =35Katthesurface( = ), θ has this value at all heights. At the trooause, the temerature is equal to the stratosheric value, 8 C =193K, and its otential temerature must also be θ =35K. Since θ = T (/ ) κ,whereκ =2/7 for the atmoshere, the ressure at the trooause must therefore be µ 1 µ 7 T κ = =1 =21.6 hpa. θ Comare the dry-adiabatic lase rate on Juiter with that of Earth given that the gravitational acceleration on Juiter is 26 m s 2 and its atmoshere is comosed almost entirely of hydrogen and therefore has a different value of c. The c of hydrogen gas is J kg 1 K 1 andsotheratioof dry-adiabatic lase rates on Juiter and Earth is, using Eq.(4.14): c Juiter c Earth = g Juiter g Earth c Earth c Juiter = = In Section 3.3 we showed that the ressure of an isothermal atmoshere varies exonentially with height. Consider now an atmoshere with uniform otential temerature. Find how ressure varies with height, and show in articular that such an atmoshere has a discrete to (where ) at altitude RT o / (κg), wherer, κ, andg have their usual meanings, and T is the temerature at 1hPa ressure. In hydrostatic balance, and for a erfect gas, z = gρ = g RT. 9

10 An atmoshere with uniform otential temerature has µ κ θ = T = constant = T, where T is the temerature where = (1hPa). Hence T = T (/ ) κ,so z = g κ 1 κ. RT Hence κ 1 d = gκ dz, RT so µ κ = κg z + C RT where C is a constant. Since z =where =, C =1and so = 1 κgz 1 κ. RT Thus, ressure vanishes at a height z = RT /κg. Note that z is a factor 1/κ (= 7/2) times the scale height based on surface temerature. Above z, the above solution is not valid (since there is no atmoshere there). Note also that T there; in fact z is exactly the height at which T following the adiabatic lase rate g/c from the surface. 11. Consider the convective circulation shown below. Air rises in the center of the system; condensation occurs at altitude z B =1km ( B =88hPa), and the convective cell (cloud is shown by 1

11 the shading) extends u to z T =9km ( T =33hPa), at which oint the air diverges and descends adiabatically in the downdraft region. The temerature at the condensation level, T B,is2 o C. Assume all condensate falls out immediately as rain. (a) Determine the secific humidity at an altitude of 3km within the cloud. (b) The uward flux of air, er unit horizontal area, through the cloud at any level z is w(z)ρ(z), whereρ is the density of dry air and w the vertical velocity. Mass balance requires that this flux be indeendent of height within the cloud. Consider the net uward flux of water vaor within the cloud and hence show that the rainfall rate below the cloud (in units of mass er unit area er unit time) is w B ρ B [q B q T ], where the subscrits B and T denote the values at cloud base and cloud to, resectively. If w B =5cm s 1, and ρ B =1.kg m 3, determine the rainfall rate in cm er day. a. Within the cloud, assume that air is saturated and the temerature therefore decreases with height at the saturated lase rate 7K km 1. At 3km altitude, 2km above cloud base, the temerature therefore is 2 14 = 6 C. At 6 C, from Fig. 1.5, saturation vaor ressure is e s ' 1hPa. To estimate ressure at this altitude, use z = gρ = g RT, andusethemeantemerature T =13 C = 286K for the layer between the level of interest and cloud base. (Since T varies by no more than ±7K from this value, the error will be no more than 2.5%.) Then ln (3km) δ (ln ) = = gδz ln B R T where δz =2km, so µ (3km) = B ex gδz = 88 ex R T µ =693hPa. 11

12 Now, since the air is saturated, its secific humidity equals q, which is q = R e s R v where R v is the gas constant for water vaor, so the secific humidity at 3km is q ' =9. gkg 1. b. The mass flux, M(z) =wρ, at altitude z, exresses the fact that a mass M of air crosses a unit cross-sectional area at altitude z er unit time. If the mixing ratio of water vaor (secific humidity) is q, then unit mass of air contains a mass q of water vaor. Therefore, the mass of water vaor crossing unit area er unit time is just Mq = wρq. Since the air is saturated everywhere within the cloud, the flux of water vaor within it is just wρq. Given that there is no mass flux out of the sides of the cloud, the net flux of water vaor into the cloud is w B ρ B q B w T ρ T q T. However, the mass flux of dry air must (because of air mass conservation) be the same at all heights, so w T ρ T = w B ρ B,whence the net flux of water vaor into the cloud (mass of water vaor er unit horizontal area er unit time) is F = w B ρ B (q B q T ). This will be ositive (since q will decrease with altitude through thecloud),sothereisanetflux of water vaor into the cloud, which must condense into liquid water. If all liquid water falls out as rain, then the rainfall rate, in terms of mass of liquid water er unit horizontal area er unit time, is just F. Within the cloud, the air is saturated, so T/ z = 7K km 1. The temerature at cloud to, 8km above cloud base, is therefore 2 56 = 36 C. From Fig. 1.5, we estimate the saturation vaor ressures as e sb = e s (2C) ' 24hPa; e st = e s ( 36C) < 1hPa. Since saturation secific humidity is q = R R v e s, 12

13 we have q B = =.169 (= 16.9 gkg 1 ) and q T < =.19 (= 1.9 gkg 1 ). Thus, neglecting q T will incur a maximum error of a little over 1%. So the net rainfall rate is r = w B ρ B [q B q T ] ' w B ρ B q B = = kg m 2 s 1. To convert this to more conventional units of deth er unit time, note that 1kg m 2 corresonds to a deth of 1/ρ w =1 3 mof water, where ρ w =1 3 kg m 3 is the density of liquid water. So r = = ms 1 ρ w = = mday 1 = 7.3 cm day Observations show that, over the Sahara, air continuously subsides (hence the Saharan climate). Consider an air arcel subsiding in this region, where the environmental temerature T e decreases with altitude at the constant rate of 7K km 1. (a) Suose the air arcel leaves height z with the environmental temerature. Assuming the dislacement to be adiabatic, show that, after a time δt, the arcel is warmer than its environment by an amount ws δt, where w is the subsidence velocity and S = dt e dz + g c, where c is the secific heatatconstantressure. 13

14 (b) Suose now that the dislacement is not adiabatic, but that the arcel cools radiatively at such a rate that its temerature is always the same as its environment (so the circulation is in equilibrium). Show that the radiative rate of energy loss er unit volume must be ρc ws, and hence that the net radiative loss to sace, er unit horizontal area, must be Z ρc ws dz = c g Z s ws d, where s is surface ressure and ρ is the air density. (c) Radiative measurements show that, over the Sahara, energy is being lost to sace at a net, annually-averaged rate of 2Wm 2.Estimate the vertically-averaged (and annually-averaged) subsidence velocity. (a) As the arcel descends adiabatically from height z (starting with the environmental temerature T e (z)), its temerature will increase according to the dry adiabatic lase rate Γ d = g/c (using our usual notation). [Note that, since the arcel will warm on descent, it will not saturate even if it is saturated initially so the dry value is the aroriate one.] After a time δt, the arcel will have descended a distance δz = wδt,wherew is the subsidence velocity, and so its temerature will be T arcel = T e (z)+γ d δz = T e (z)+ g c wδt. The environmental temerature at height z δz (given that the environmental lase rate is linear in z) is T e (z δz) =T e (z) dt e dz δz = T e(z) dt e dz wδt. Therefore the temerature excess of the arcel is δt = T arcel T e (z δz) =Sw δt, where S = dt e dz + g c. 14

15 Note that S is a measure of the static stability of the environment it exresses the dearture of the atmosheric lase rate from the adiabatic value. Since we are told that dt e /dz = 7K km 1 is constant, S will be constant also, S =3Kkm 1. (b) In order for the arcel to maintain the environmental temerature, it must cool. Think of the sequence as a series of stes: adiabatic descent for a very short time δt, followed by non-adiabatic cooling to the environmental temerature (at constant ressure, since the arcel ressure must always equal that of the environment). After descending for a short time δt, the required heat loss er unit volume, δq, is δq = ρc δt = ρc ws δt. Therefore the heat loss er unit volume er unit time is δq δt = ρc ws. Integrating in the vertical, and then using hydrostatic balance d = gρ dz, the net cooling er unit horizontal area is J cool = Z δq δt dz = Z ρc ws dz = c g Z s ws d. (3) QED. Note that the greater the stability S, or subsidence velocity w, the greater the cooling must be to sustain equilibrium. (c) Observations show that, on the annual average, J cool =2Wm 2. Since S is constant, we can rewrite (3) as J cool = c S g s hwi, where hwi = 1 Z s wd s is the vertically (ressure) averaged and annually-averaged subsidence velocity. Therefore hwi = gj cool c S s = ms 1 =.65 mm s 1. 15