444 QQMZ - University of British Columbia · following radius of curvature (km). Assume the air...

ATSC201Fall2019Assignment12AnswerKeyTotalmarksoutof50Chapter13:A3i,A7i,A14i,A16iChapter17:A6i,A8i,A12i,A13i,A17i,A20a,A35i

Chapter13

A3i)

Given: Δz_mtn= 2 kmΔz_T= 12 kmφ= 45 deg

Find: A= ? km

Useeqn13.3:

wherefc/β=Rearth*tan(φ)frompage444beloweqn13.3Rearth= 6371 km

fc/β= 6371 km

A= 1061.83333 km

Check:Unitsok.Physicsok.Discussion:Thehigherthemountains,theshorterthecolumnofairwillbe,andwiththisgreaterchange,relativevorticitywillneedtodecreasemoretoconservepotentialvorticity.ThisdecreasingrelativevorticityiswhatinitiatestheRossbywave.

474 CHAPTER 13 • EXTRATROPICAL CYCLONES

“C” in Fig. 13.20 given an average wind speed (m s–1) of: a. 20 b. 25 c. 30 d. 35 e. 40 f. 45 g. 50 h. 55 i. 60 j. 65 k. 70 l. 75 m. 80 n. 85

A8. Regarding equatorward propagation of cy-clones on the eastern slope of mountains, if a cyclone of radius 1000 km and potential vorticity of 3x10–8 m–1s–1 is over a slope as given below (∆z/∆x), find the change in relative vorticity (s–1) between the north and south sides of the cyclone. a. 1/500 b. 1/750 c. 1/1000 d. 1/1250 e. 1/1500 f. 1/1750 g. 1/2000 h. 1/2250 i. 1/2500 j. 1/2750 k. 1/3000 l. 1/3250

A9. Recall from the Atmospheric Forces and Winds chapter that incompressible mass continuity implies that ∆W/∆z = –D, where D is horizontal divergence. Find the Coriolis contribution to the stretching term (s–2) in the relative-vorticity tendency equation, giv-en the average 85 kPa divergence in Fig. 13.26 for the following USA state: a. AZ b. WI c. KS d. KY e. VA f. CO g. KA h. AB i. MD j. ID k. central TX l. NY m. IN

A10. Find the spin-up rate (s–2) of quasi-geostrophic vorticity, assuming that the following is the only non-zero characteristic: a. Geostrophic wind of 30 m s–1 from north with-in region where geostrophic vorticity increases to-ward the north by 6x10–5 s–1 over 500 km distance. b. Geostrophic wind of 50 m s–1 from west with-in region where geostrophic vorticity increases to-ward the east by 8x10–5 s–1 over 1000 km distance. c. A location at 40°N with geostrophic wind from the north of 25 m s–1. d. A location at 50°N with geostrophic wind from the south of 45 m s–1. e. A location at 35°N with vertical velocity increas-ing 0.5 m s–1 with each 1 km increase in height. f. A location at 55°N with vertical velocity decreas-ing 0.2 m s–1 with each 2 km increase in height.

A11. Find the value of geostrophic vorticity (s–1), given the following changes of (Ug, Vg) in m s–1 with 500 km distance toward the (north, east): a. (0, 5) b. (0, 10) c. (0, –8) d. (0, –20) e. (7, 0) f. (15, 0) g. (–12, 0) h. (-25, 0) i. (5, 10) j. (20, 10) k. (–10, 15) l. (–15, –12)

A12. Find the value of geostrophic vorticity (s–1), given a geostrophic wind speed of 35 m s–1 with the following radius of curvature (km). Assume the air rotates similar to a solid-body. a. 450 b. –580 c. 690 d. –750 e. 825 f. –988 g. 1300 h. –1400 i. 2430 j. –2643 k. 2810 l. –2900 m. 3014 n. –3333

A1. For latitude 50°N, find the approximate wave-length (km) of upper-atmosphere (Rossby) waves triggered by mountains, given an average wind speed (m s–1) of: a. 20 b. 25 c. 30 d. 35 e. 40 f. 45 g. 50 h. 55 i. 60 j. 65 k. 70 l. 75 m. 80 n. 85

A2. Find the rate of increase of the parameter (i.e., the rate of change of Coriolis parameter with dis-tance north) in units of m–1s–1 at the following lati-tude (°N): a. 40 b. 45 c. 50 d. 55 e. 60 f. 65 g. 70 h. 80 i. 35 j. 30 k. 25 l. 20 m. 15 n. 10

A3. Given a tropospheric depth of 12 km at latitude 45°N, what is the meridional (north-south) ampli-tude (km) of upper-atmosphere (Rossby) waves trig-gered by mountains, given an average mountain-range height (km) of: a. 0.4 b. 0.6 c. 0.8 d. 1.0 e. 1.2 f. 1.4 g. 1.6 h. 1.8 i. 2.0 j. 2.2 k. 2.4 l. 2.6 m. 2.8 n. 3.0

A4. How good is the approximation of eq. (13.4) to eq. (13.3) at the following latitude (°N)? a. 40 b. 45 c. 50 d. 55 e. 60 f. 65 g. 70 h. 80 i. 35 j. 30 k. 25 l. 20 m. 15 n. 10

A5. For a troposphere of depth 12 km at latitude 43°N, find the potential vorticity in units of m–1s–1, given the following: [wind speed (m s–1) , radius of curvature (km)] a. 50 , 500 b. 50 , 1000 c. 50 , 1500 d. 50 , 2000 e. 50 , –500 f. 50 , – 1000 g. 50 , –1500 h. 50 , -2000 i. 75 , 500 j. 75 , 1000 k. 75 , 1500 l. 75 , 2000 m. 75 , –1000 n. 75 , – 2000

A6. For air at 55°N with initially no curvature, find the potential vorticity in units of m–1s–1 for a tropo-sphere of depth (km): a. 7.0 b. 7.5 c. 8.0 d. 8.5 e. 9.0 f. 9.5 g. 10.0 h. 10.5 i. 11 j. 11.5 k. 12 l. 12.5 m. 13 n. 13.5

A7. When air at latitude 60°N flows over a mountain range of height 2 km within a troposphere of depth 12 km, find the radius of curvature (km) at location


Consider a wind that causes air in the troposphere to blow over the Rocky mountains (Fig. 13.20). Con-vective clouds (e.g., thunderstorms) and turbulence can cause the Rossby wave amplitude to decrease further east, so the first wave after the mountain (at location c in Fig. 13.20) is the one you should focus on. These Rossby waves have a dominant wave-length ( ) of roughly

21 2

· ·/

M (13.1)

where the mean wind speed is M. As you have seen in an earlier chapter, is the northward gradient of the Coriolis parameter ( fc):

fy Rc

earth

2 ·· cos •(13.2)

Factor 2· = 1.458x10–4 s–1 is twice the angular rota-tion rate of the Earth. At North-American latitudes,

is roughly 1.5 to 2x10–11 m–1 s–1. Knowing the mountain-range height (∆zmtn) rela-tive to the surrounding plains, and knowing the ini-tial depth of the troposphere (∆zT), the Rossby-wave amplitude A is:

Af z

zc mtn

T· (13.3)

Because is related to fc, we can analytically find their ratio as fc/ = REarth·tan( ), where the average radius (REarth) of the Earth is 6371 km. Over North America the tangent of the latitude is tan( ) ≈ 1. Thus:

Azz

Rmtn

Tearth· (13.4)

where 2A is the ∆y distance between the wave trough and crest. In summary, the equations above show that north-south Rossby-wave amplitude depends on the height of the mountains, but does not depend on wind speed. Conversely, wind speed is important in determining Rossby wavelength, while mountain height is irrelevant.

Use conservation of potential vorticity as a tool to understand such mountain lee-side Rossby-wave triggering (Fig. 13.20). Create a “toy model” by as-suming wind speed is constant in the Rossby wave,

Sample Application What amplitude & wavelength of terrain-triggered Rossby wave would you expect for a mountain range at 48°N that is 1.2 km high? The upstream depth of the troposphere is 11 km, with upstream wind is 19 m s–1.

Find the AnswerGiven: = 48°N, ∆zmtn = 1.2 km, ∆zT = 11 km, M = 19 m s–1.Find: A = ? km , = ? km

Use eq. (13.4): A = [ (1.2 km) / (11 km) ] · (6371 km) = 695 km

Next, use eq. (13.2) to find at 48°N: = (2.294x10–11 m–1·s–1) · cos(48°) = 1.53x10–11 m–1·s–1

Finally, use eq. (13.1):

219

1 53 10 11

1 2

· ··

. ·

/m s

m s

1

1 1 = 6990 km

Check: Physics and units are reasonable.Exposition: Is this wave truly a planetary wave? Yes, because its wavelength (6,990 km) would fit 3.8 times around the Earth at 48°N (circumference = 2·π·REarth·cos(48°) = 26,785 km). Also, the north-south meander of the wave spans 2A = 12.5° of latitude.

Figure 13.20Cyclogenesis to the lee of the mountains. (a) Vertical cross sec-tion. (b) Map of jet-stream flow. “Ridge” and “trough” refer to the wind-flow pattern, not the topography.

A7i)

Given: φ= 60 ° 1.04719755 radiansΔz_mtn= 2 km 2000 mΔz_T= 12 km 12000 mM= 60 m/s 0.06 km/s

Find: Rc= ? km

CanassumethattheridgeafterlocationcisatthesamelatitudeaslocationaThereforecanuseequation13.4tofindtheamplitudeoftheRossbywave.

whereR_earth(km)= 6371(m)= 6371000

A= 1061.83 km 2A= 2123.66667 km1061833.33 m

111km=1°lat 111so2Aindegs.Lat= 19.1321321 °Lat

soφ(c)= φ(a)-φ(2A)φ(c)= 40.87 °N 0.71327885 radians

ThenfindFcataandc

useeqn10.16 where2Ω= 1.46E-04 1/s






























Consider a wind that causes air in the troposphere to blow over the Rocky mountains (Fig. 13.20). Con-vective clouds (e.g., thunderstorms) and turbulence can cause the Rossby wave amplitude to decrease further east, so the first wave after the mountain (at location c in Fig. 13.20) is the one you should focus on. These Rossby waves have a dominant wave-length ( ) of roughly

21 2

· ·/

M (13.1)

where the mean wind speed is M. As you have seen in an earlier chapter, is the northward gradient of the Coriolis parameter ( fc):

fy Rc

earth

2 ·· cos •(13.2)

Factor 2· = 1.458x10–4 s–1 is twice the angular rota-tion rate of the Earth. At North-American latitudes,

is roughly 1.5 to 2x10–11 m–1 s–1. Knowing the mountain-range height (∆zmtn) rela-tive to the surrounding plains, and knowing the ini-tial depth of the troposphere (∆zT), the Rossby-wave amplitude A is:

Af z

zc mtn

T· (13.3)

Because is related to fc, we can analytically find their ratio as fc/ = REarth·tan( ), where the average radius (REarth) of the Earth is 6371 km. Over North America the tangent of the latitude is tan( ) ≈ 1. Thus:

Azz

Rmtn

Tearth· (13.4)

where 2A is the ∆y distance between the wave trough and crest. In summary, the equations above show that north-south Rossby-wave amplitude depends on the height of the mountains, but does not depend on wind speed. Conversely, wind speed is important in determining Rossby wavelength, while mountain height is irrelevant.

Use conservation of potential vorticity as a tool to understand such mountain lee-side Rossby-wave triggering (Fig. 13.20). Create a “toy model” by as-suming wind speed is constant in the Rossby wave,

Sample Application What amplitude & wavelength of terrain-triggered Rossby wave would you expect for a mountain range at 48°N that is 1.2 km high? The upstream depth of the troposphere is 11 km, with upstream wind is 19 m s–1.

Find the AnswerGiven: = 48°N, ∆zmtn = 1.2 km, ∆zT = 11 km, M = 19 m s–1.Find: A = ? km , = ? km

Use eq. (13.4): A = [ (1.2 km) / (11 km) ] · (6371 km) = 695 km

Next, use eq. (13.2) to find at 48°N: = (2.294x10–11 m–1·s–1) · cos(48°) = 1.53x10–11 m–1·s–1

Finally, use eq. (13.1):

219

1 53 10 11

1 2

· ··

. ·

/m s

m s

1

1 1 = 6990 km

Check: Physics and units are reasonable.Exposition: Is this wave truly a planetary wave? Yes, because its wavelength (6,990 km) would fit 3.8 times around the Earth at 48°N (circumference = 2·π·REarth·cos(48°) = 26,785 km). Also, the north-south meander of the wave spans 2A = 12.5° of latitude.

Figure 13.20Cyclogenesis to the lee of the mountains. (a) Vertical cross sec-tion. (b) Map of jet-stream flow. “Ridge” and “trough” refer to the wind-flow pattern, not the topography.

fc=2*Ω*sin(φ)

fc(atc)= 9.54E-05 1/sfc(ata)= 1.26E-04 1/s

Useeqn13.6and13.5

KnowingthatpotentialvorticityisconservedandthatΔzT(ata)=ΔzT(atc)wecanrearragethesetobe:

R(atc)=M/(fc(ata)-fc(atc))

R(atc)= 1943816.47 m1943.82 km

Check:Unitsok.Physicsok.Discussion: Theradiusofcurvatureatlocationcmustbepositivebecauseitisturningcyclonicallytokeeppotentialvorticityconstant.Theradiusofcurvatureequationatlocationb(13.7)doesnotapplyatlocationcbecausetheCoriolisforceissmaller(flowisclosertoequator).

A14i)

Given: P= 85 kPaW= -0.5 m/s

Find: ω= ? Pa/s

Useeqn.13.14:

FromTable1-5estimateρρ= 1.07 kg/m^3 |g|= 9.8 m/s^2

R. STULL • PRACTICAL METEOROLOGY 475

A13. What is the value of omega (Pa s–1) following a vertically-moving air parcel, if during 1 minute its pressure change (kPa) is: a. –2 b. –4 c. –6 d. –8 e. –10 f. –12 g. –14 h. –16 i. –18 j. –20 k. 0.00005 l. 0.0004 m. 0.003 n. 0.02

A14. At an altitude where the ambient pressure is 85 kPa, convert the following vertical velocities (m s–1) into omega (Pa s–1): a. 2 b. 5 c. 10 d. 20 e. 30 f. 40 g. 50 h. –0.2 i. –0.5 j. –1.0 k. –3 l. –5 m. – 0.03

A15. Using Fig. 13.15, find the most extreme value horizontal divergence (10–5 s–1) at 20 kPa over the following USA state: a. MI b. WI c. IL d. IN e. TN f. GA g. MS h. AB i. KY j. PA k. NY l. SC

A16. Find the vertical velocity (m s–1) at altitude 9 km in an 11 km thick troposphere, if the divergence (10–5 s–1) given below occurs within a 2 km thick layer within the top of the troposphere. a. 0.2 b. 1 c. 1.5 d. 2 e. 3 f. 4 g. 5 h. 6 i. –0.3 j.–0.7 k. –1.8 l. –2.2 m. –3.5 n. –5

A17. Jet-stream inflow is 30 m s–1 in a 4 km thick lay-er near the top of the troposphere. Jet-stream out-flow (m s–1) given below occurs 800 km downwind within the same layer. Find the vertical velocity (m s–1) at the bottom of that layer a. 35 b. 40 c. 45 d. 50 e. 55 f. 60 g. 65 h. 70 i. 30 j. 25 k. 20 l. 15 m. 10 n. 5

A18. Find the diagonal distance (km) from trough to crest in a jet stream for a wave of 750 km amplitude with wavelength (km) of: a. 1000 b. 1300 c. 1600 d. 2000 e. 2200 f. 2500 g. 2700 h. 3000 i. 3100 j. 3300 k. 3800 l. 4100 m. 4200 n. 4500

A19. Given the data from the previous exercise, find the radius (km) of curvature near the crests of a si-nusoidal wave in the jet stream.

A20. Find the gradient-wind speed difference (m s–1) between the jet-stream speed moving through the anticyclonic crest of a Rossby wave in the N. Hemi-sphere and the jet-stream speed moving through the trough. Use data from the previous 2 exercises, and assume a geostrophic wind speed of 75 m s–1 for a wave centered on latitude 40°N.

A21. Given the data from the previous 3 exercises. Assuming that the gradient-wind speed difference

calculated in the previous exercise is valid over a layer between altitudes 8 km and 12 km, where the tropopause is at 12 km, find the vertical velocity (m s–1) at 8 km altitude.

A22. Suppose that a west wind enters a region at the first speed (m s–1) given below, and leaves 500 km downwind at the second speed (m s–1). Find the north-south component of ageostrophic wind (m s–1) in this region. Location is 55°N. a. (40, 50) b. (30, 60) c. (80, 40) d. (70, 50) e. (40, 80) f. (60, 30) g. (50, 40) h. (70, 30) i. (30, 80) j. (40, 70) k. (30, 70) l. (60, 20)

A23. Use the ageostrophic right-hand rule to find the ageostrophic wind direction for the data of the previous problem.

A24. Using the data from A22, find the updraft speed (m s–1) into a 4 km thick layer at the top of the troposphere, assuming the half-width of the jet streak is 200 km.

A25. Suppose that the thickness of the 100 - 50 kPa layer is 5.5 km and the Coriolis parameter is 10–4 s–1. A 20 m s–1 thermal wind from the west blows across a domain of x dimension given below in km. Across that domain in the x-direction is a decrease of cy-clonic relative vorticity of 3x10–4 s–1. What is the value of mid-tropospheric ascent velocity (m s–1), based on the omega equation? a. 200 b. 300 c. 400 d. 500 e. 600 f. 700 g. 800 h. 900 i. 1000 j. 1100 k. 1200 l. 1300 m. 1400 n. 1400

A26. On the 70 kPa isobaric surface, ∆Ug/∆x = (4 m s–1)/(500 km) and ∆T/∆x = ___°C/(500 km), where the temperature change is given below. All other gradi-ents are zero. Find the Q-vector components Qx, Qy, and the magnitude and direction of Q. a. 1 b. 1.5 c. 2 d. 2.5 e. 3 f. 3.5 g. 4 h. 4.5 i. 5 j. 5.5 k. 6 l. 6.5 m. 7 n. 7.5

A27. Find Q-vector magnitude on the 85 kPa isobaric surface if the magnitude of the horizontal tempera-ture gradient is 5°C/200 km, and the magnitude of the geostrophic-wind difference-vector component (m s–1) along an isotherm is __ /200 km, where __ is: a. 1 b. 1.5 c. 2 d. 2.5 e. 3 f. 3.5 g. 4 h. 4.5 i. 5 j. 5.5 k. 6 l. 6.5 m. 7 n. 7.5

A28. Find the mass of air over 1 m2 of the Earth’s surface if the surface pressure (kPa) is: a. 103 b. 102 c. 101 d. 99 e. 98 f. 97 g. 96 h. 95 i. 93 j. 90 k. 85 l. 80 m. 75 n. 708


You can also use updraft speed to estimate cy-clone strength and the associated clouds. For the case-study cyclone, Fig. 13.25 shows upward motion near the middle of the troposphere (at 50 kPa). Recall from classical physics that the definition of vertical motion is W = ∆z/∆t, for altitude z and time t. Because each altitude has an associated pressure, define a new type of vertical velocity in terms of pressure. This is called omega ( ):

Pt •(13.13)

Omega has units of Pa s–1. You can use the hypsometric equation to relate W and : · ·g W •(13.14)

for gravitational acceleration magnitude |g| = 9.8 m·s–2 and air density . The negative sign in eq. (13.14) implies that updrafts (positive W) are associ-ated with negative . As an example, if your weath-er map shows = –0.68 Pa s–1 on the 50 kPa surface, then the equation above can be rearranged to give W = 0.1 m s–1, where an mid-tropospheric density of ≈ 0.69 kg·m–3 was used. Use either W or to represent vertical motion. Nu-merical weather forecasts usually output the vertical velocity as . For example, Fig. 13.28 shows upward motion ( ) near the middle of the troposphere (at 50 kPa). The following three methods will be employed to study ascent: the continuity equation, the omega equation, and Q-vectors. Near the tropopause, hori-zontal divergence of jet-stream winds can force mid-tropospheric ascent in order to conserve air mass as given by the continuity equation. The almost-geo-strophic (quasi-geostrophic) nature of lower-tropo-spheric winds allows you to estimate ascent at 50 kPa using thermal-wind and vorticity principles in the omega equation. Q-vectors consider ageostrophic motions that help maintain quasi-geostrophic flow. These methods are just different ways of looking at the same processes, and they often give similar re-sults.

Sample Application At an elevation of 5 km MSL, suppose (a) a thun-derstorm has an updraft velocity of 40 m s–1, and (b) the subsidence velocity in the middle of an anticyclone is –0.01 m s–1. Find the corresponding omega values.

Find the AnswerGiven: (a) W = 40 m s–1. (b) W = – 0.01 m s–1. z = 5 km.Find: = ? kPa s –1 for (a) and (b).

To estimate air density, use the standard atmosphere table from Chapter 1: = 0.7361 kg m–3 at z = 5 km.

Next, use eq. (13.14) to solve for the omega values:(a) = –(0.7361 kg m–3)·(9.8 m s–2)·(40 m s–1) = –288.55 (kg·m–1·s–2)/s = –288.55 Pa s–1 = –0.29 kPa/s

(b) = –(0.7361 kg m–3)·(9.8 m s–2)·(–0.01 m s–1) = 0.0721 (kg·m–1·s–2)/s = 0.0721 Pa s–1 = 7.21x10–5 kPa s–1

Check: Units and sign are reasonable.Exposition: CAUTION. Remember that the sign of omega is opposite that of vertical velocity, because as height increases in the atmosphere, the pressure de-creases. As a quick rule of thumb, near the surface where air density is greater, omega (in kPa s–1) has magnitude of roughly a hundredth of W (in m s–1), with opposite sign.

Figure 13.28Vertical velocity (omega) in pressure coordinates, for the case-study cyclone. Negative omega (colored red on this map) cor-responds to updrafts.

50 kPa Omega (Pa/s). Red up. Valid 12 UTC, 4 Apr 2014

X


and that there is no wind shear affecting vorticity. For this situation, the conservation of potential vorticity p is given by eq. (11.25) as:

pcM R f

z( / )

constant •(13.5)

For this toy model, consider the initial winds to be blowing straight toward the Rocky Mountains from the west. These initial winds have no curvature at location “a”, thus R = ∞ and eq. (13.5) becomes:

p

c a

T a

fz

.

.

(13.6)

where ∆zT.a is the average depth of troposphere at point “a”. Because potential vorticity is conserved, we can use this fixed value of p to see how the Ross-by wave is generated. Let ∆zmtn be the relative mountain height above the surrounding land (Fig. 13.20a). As the air blows over the mountain range, the troposphere becomes thinner as it is squeezed between mountain top and the tropopause at location “b”: ∆zT.b = ∆zT.a – ∆zmtn. But the latitude of the air hasn’t changed much yet, so fc.b ≈ fc.a. Because ∆z has changed, we can solve eq. (13.5) for the radius of curvature needed to main-tain p.b = p.a.

RM

f z zbc a mtn T a. .·( / ) (13.7)

Namely, in eq. (13.5), when ∆z became smaller while fc was constant, M/R had to also become smaller to keep the ratio constant. But since M/R was initially zero, the new M/R had to become negative. Nega-tive R means anticyclonic curvature. As sketched in Fig. 13.20, such curvature turns the wind toward the equator. But equatorward-moving air experiences smaller Coriolis parameter, requiring that Rb become larger (less curved) to con-serve p . Near the east side of the Rocky Mountains the terrain elevation decreases at point “c”, allowing the air thickness ∆z to increase back to is original value. But now the air is closer to the equator where Coriolis parameter is smaller, so the radius of cur-vature Rc at location “c” becomes positive in order to keep potential vorticity constant. This positive vorticity gives that cyclonic curvature that defines the lee trough of the Rossby wave. As was sketched in Fig. 13.7, surface cyclogenesis could be supported just east of the lee trough.

Sample Application Picture a scenario as plotted in Fig. 13.20, with 25 m s–1 wind at location “a”, mountain height of 1.2 km, tro-posphere thickness of 11 km, and latitude 45°N. What is the value of the initial potential vorticity, and what is the radius of curvature at point “b”?

Find the AnswerGiven: M = 25 m s–1, ∆zmtn = 1.2 km, Rinitial = ∞, ∆zT = 11 km, = 45°N. Find: p.a = ? m–1·s–1, Rb = ? km

Assumption: Neglect wind shear in the vorticity cal-culation.

Eq. (10.16) can be applied to get the Coriolis parameter fc = (1.458x10–4 s–1)·sin(45°) = 1.031x10–4 s–1

Use eq. (13.6):

p1 031 10

11

4. skm

1 = 9.37x10–9 m–1 s–1

Next, apply eq. (13.7) to get the radius of curvature:

Rb( )

( . )·( . / )25

1 031 10 1 2 114m/s

s km km1 = –2223. km

Check: Physics and units are reasonable.Exposition: The negative sign for the radius of cur-vature means that the turn is anticyclonic (clockwise in the N. Hemisphere). Typically, the cyclonic trough curvature is the same order of magnitude as the anti-cyclonic ridge curvature. East of the first trough and west of the next ridge is where cyclogenesis is sup-ported.


and that there is no wind shear affecting vorticity. For this situation, the conservation of potential vorticity p is given by eq. (11.25) as:

pcM R f

z( / )

constant •(13.5)

For this toy model, consider the initial winds to be blowing straight toward the Rocky Mountains from the west. These initial winds have no curvature at location “a”, thus R = ∞ and eq. (13.5) becomes:

p

c a

T a

fz

.

.

(13.6)

where ∆zT.a is the average depth of troposphere at point “a”. Because potential vorticity is conserved, we can use this fixed value of p to see how the Ross-by wave is generated. Let ∆zmtn be the relative mountain height above the surrounding land (Fig. 13.20a). As the air blows over the mountain range, the troposphere becomes thinner as it is squeezed between mountain top and the tropopause at location “b”: ∆zT.b = ∆zT.a – ∆zmtn. But the latitude of the air hasn’t changed much yet, so fc.b ≈ fc.a. Because ∆z has changed, we can solve eq. (13.5) for the radius of curvature needed to main-tain p.b = p.a.

RM

f z zbc a mtn T a. .·( / ) (13.7)

Namely, in eq. (13.5), when ∆z became smaller while fc was constant, M/R had to also become smaller to keep the ratio constant. But since M/R was initially zero, the new M/R had to become negative. Nega-tive R means anticyclonic curvature. As sketched in Fig. 13.20, such curvature turns the wind toward the equator. But equatorward-moving air experiences smaller Coriolis parameter, requiring that Rb become larger (less curved) to con-serve p . Near the east side of the Rocky Mountains the terrain elevation decreases at point “c”, allowing the air thickness ∆z to increase back to is original value. But now the air is closer to the equator where Coriolis parameter is smaller, so the radius of cur-vature Rc at location “c” becomes positive in order to keep potential vorticity constant. This positive vorticity gives that cyclonic curvature that defines the lee trough of the Rossby wave. As was sketched in Fig. 13.7, surface cyclogenesis could be supported just east of the lee trough.

Sample Application Picture a scenario as plotted in Fig. 13.20, with 25 m s–1 wind at location “a”, mountain height of 1.2 km, tro-posphere thickness of 11 km, and latitude 45°N. What is the value of the initial potential vorticity, and what is the radius of curvature at point “b”?

Find the AnswerGiven: M = 25 m s–1, ∆zmtn = 1.2 km, Rinitial = ∞, ∆zT = 11 km, = 45°N. Find: p.a = ? m–1·s–1, Rb = ? km

Assumption: Neglect wind shear in the vorticity cal-culation.

Eq. (10.16) can be applied to get the Coriolis parameter fc = (1.458x10–4 s–1)·sin(45°) = 1.031x10–4 s–1

Use eq. (13.6):

p1 031 10

11

4. skm

1 = 9.37x10–9 m–1 s–1

Next, apply eq. (13.7) to get the radius of curvature:

Rb( )

( . )·( . / )25

1 031 10 1 2 114m/s

s km km1 = –2223. km

Check: Physics and units are reasonable.Exposition: The negative sign for the radius of cur-vature means that the turn is anticyclonic (clockwise in the N. Hemisphere). Typically, the cyclonic trough curvature is the same order of magnitude as the anti-cyclonic ridge curvature. East of the first trough and west of the next ridge is where cyclogenesis is sup-ported.

ω= 5.243 Pa/s

Check:Unitsok.Physicsok.Discussion:Thetermomega,bydefinition,isthechangeofpressurewithtime.So,whenomegaisnegative,thismeansairisdescendingandthatpressureisincreasingastheairmoves

A16h)

Given: D= -3.00E-06 /sΔz= 2 km 2000 m

Find: Wmid= ? m/s

Useeqn.13.15:

Wmid= -0.01 m/s

Check:Unitsok.Physicsok.Discussion:Duetotheconservationofmass(continuityequation)ifthereishorizontaldivergence,thentheremustbeverticalmotiontofillintheairthatisleavinghorizontally.

Chapter17

A6i)

Given: α= 50 °CD= 0.002h= 5 mTve= 20 °C 293 K


A13. What is the value of omega (Pa s–1) following a vertically-moving air parcel, if during 1 minute its pressure change (kPa) is: a. –2 b. –4 c. –6 d. –8 e. –10 f. –12 g. –14 h. –16 i. –18 j. –20 k. 0.00005 l. 0.0004 m. 0.003 n. 0.02

A14. At an altitude where the ambient pressure is 85 kPa, convert the following vertical velocities (m s–1) into omega (Pa s–1): a. 2 b. 5 c. 10 d. 20 e. 30 f. 40 g. 50 h. –0.2 i. –0.5 j. –1.0 k. –3 l. –5 m. – 0.03

A15. Using Fig. 13.15, find the most extreme value horizontal divergence (10–5 s–1) at 20 kPa over the following USA state: a. MI b. WI c. IL d. IN e. TN f. GA g. MS h. AB i. KY j. PA k. NY l. SC

A16. Find the vertical velocity (m s–1) at altitude 9 km in an 11 km thick troposphere, if the divergence (10–5 s–1) given below occurs within a 2 km thick layer within the top of the troposphere. a. 0.2 b. 1 c. 1.5 d. 2 e. 3 f. 4 g. 5 h. 6 i. –0.3 j.–0.7 k. –1.8 l. –2.2 m. –3.5 n. –5

A17. Jet-stream inflow is 30 m s–1 in a 4 km thick lay-er near the top of the troposphere. Jet-stream out-flow (m s–1) given below occurs 800 km downwind within the same layer. Find the vertical velocity (m s–1) at the bottom of that layer a. 35 b. 40 c. 45 d. 50 e. 55 f. 60 g. 65 h. 70 i. 30 j. 25 k. 20 l. 15 m. 10 n. 5

A18. Find the diagonal distance (km) from trough to crest in a jet stream for a wave of 750 km amplitude with wavelength (km) of: a. 1000 b. 1300 c. 1600 d. 2000 e. 2200 f. 2500 g. 2700 h. 3000 i. 3100 j. 3300 k. 3800 l. 4100 m. 4200 n. 4500

A19. Given the data from the previous exercise, find the radius (km) of curvature near the crests of a si-nusoidal wave in the jet stream.

A20. Find the gradient-wind speed difference (m s–1) between the jet-stream speed moving through the anticyclonic crest of a Rossby wave in the N. Hemi-sphere and the jet-stream speed moving through the trough. Use data from the previous 2 exercises, and assume a geostrophic wind speed of 75 m s–1 for a wave centered on latitude 40°N.

A21. Given the data from the previous 3 exercises. Assuming that the gradient-wind speed difference

calculated in the previous exercise is valid over a layer between altitudes 8 km and 12 km, where the tropopause is at 12 km, find the vertical velocity (m s–1) at 8 km altitude.

A22. Suppose that a west wind enters a region at the first speed (m s–1) given below, and leaves 500 km downwind at the second speed (m s–1). Find the north-south component of ageostrophic wind (m s–1) in this region. Location is 55°N. a. (40, 50) b. (30, 60) c. (80, 40) d. (70, 50) e. (40, 80) f. (60, 30) g. (50, 40) h. (70, 30) i. (30, 80) j. (40, 70) k. (30, 70) l. (60, 20)

A23. Use the ageostrophic right-hand rule to find the ageostrophic wind direction for the data of the previous problem.

A24. Using the data from A22, find the updraft speed (m s–1) into a 4 km thick layer at the top of the troposphere, assuming the half-width of the jet streak is 200 km.

A25. Suppose that the thickness of the 100 - 50 kPa layer is 5.5 km and the Coriolis parameter is 10–4 s–1. A 20 m s–1 thermal wind from the west blows across a domain of x dimension given below in km. Across that domain in the x-direction is a decrease of cy-clonic relative vorticity of 3x10–4 s–1. What is the value of mid-tropospheric ascent velocity (m s–1), based on the omega equation? a. 200 b. 300 c. 400 d. 500 e. 600 f. 700 g. 800 h. 900 i. 1000 j. 1100 k. 1200 l. 1300 m. 1400 n. 1400

A26. On the 70 kPa isobaric surface, ∆Ug/∆x = (4 m s–1)/(500 km) and ∆T/∆x = ___°C/(500 km), where the temperature change is given below. All other gradi-ents are zero. Find the Q-vector components Qx, Qy, and the magnitude and direction of Q. a. 1 b. 1.5 c. 2 d. 2.5 e. 3 f. 3.5 g. 4 h. 4.5 i. 5 j. 5.5 k. 6 l. 6.5 m. 7 n. 7.5

A27. Find Q-vector magnitude on the 85 kPa isobaric surface if the magnitude of the horizontal tempera-ture gradient is 5°C/200 km, and the magnitude of the geostrophic-wind difference-vector component (m s–1) along an isotherm is __ /200 km, where __ is: a. 1 b. 1.5 c. 2 d. 2.5 e. 3 f. 3.5 g. 4 h. 4.5 i. 5 j. 5.5 k. 6 l. 6.5 m. 7 n. 7.5

A28. Find the mass of air over 1 m2 of the Earth’s surface if the surface pressure (kPa) is: a. 103 b. 102 c. 101 d. 99 e. 98 f. 97 g. 96 h. 95 i. 93 j. 90 k. 85 l. 80 m. 75 n. 708

682 CHAPTER 17 • REGIONAL WINDS

A14. Assume |g|/Tv = 0.0333 m·s–2·K–1. Find the in-ternal wave horizontal group speed (m s–1) for a sta-bly stratified air layer of depth 400 m, given ∆ v/∆z (K km–1) of: a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A15. For the previous problem, find the value of the Froude number Fr2. Also, classify this flow as sub-critical, critical, or supercritical.

A16. Winds of 10 m s–1 are flowing in a valley of 10 km width. Further downstream, the valley narrows to the width (km) given below. Find the wind speed (m s–1) in the constriction, assuming constant depth flow. a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A17. Assume |g|/Tv = 0.0333 m·s–2·K–1. For a two-layer atmospheric system flowing through a short gap, find the maximum expected gap wind speed (m s–1). Flow depth is 300 m, and the virtual poten-tial temperature difference (K) is: a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A18. Find the long-gap geostrophic wind (m s–1) at latitude 50°, given |g|/Tv = 0.0333 m·s–2·K–1 and ∆ v = 3°C, and assuming that the slope of the top cold-air surface is given by the height change (km) below across a valley 10 km wide. a. 0.3 b. 0.4 c. 0.5 d. 0.6 e. 0.7 f. 0.8 g. 0.9 h. 1.0 i. 1.1 j. 1.2 k. 2.4 m. 2.6 n. 2.8 o. 3

A19. Find the external Rossby radius of deformation (km) for a coastally trapped jet that rides against a mountain range of 2500 m altitude at latitude (°) giv-en below, for air that is colder than its surroundings by 10°C. Assume |g|/Tv = 0.0333 m·s–2·K–1. a. 80 b. 85 c. 20 d. 25 e. 30 f. 35 g. 40 h. 45 i. 50 j. 55 k. 60 m. 65 n. 70 o. 75

A20. Assume |g|/Tv = 0.0333 m·s–2·K–1. Find the natural wavelength of air, given a. M = 2 m s–1, ∆T/∆z = 5 °C km–1

b. M = 20 m s–1, ∆T/∆z = –8 °C km–1

c. M = 5 m s–1, ∆T/∆z = –2 °C km–1

d. M = 20 m s–1, ∆T/∆z = 5 °C km–1

e. M = 5 m s–1, ∆T/∆z = –8 °C km–1

f. M = 2 m s–1, ∆T/∆z = –2 °C km–1

g. M = 5 m s–1, ∆T/∆z = 5 °C km–1

h. M = 2 m s–1, ∆T/∆z = –8 °C km–1

A4. Anabatic flow has a temperature excess of 4°C. Find the buoyant along-slope pressure gradient force per unit mass for a slope of angle (°): a. 10 b. 15 c. 20 d. 25 e. 30 f. 35 g. 40 h. 45 i. 50 j. 55 k. 60 m. 65 n. 70 o. 75

A5(§). Plot katabatic wind speed (m s–1) vs. downslope distance (m) if the environment is 20°C and the cold katabatic air is 15°C. The slope angle (°) is: a. 10 b. 15 c. 20 d. 25 e. 30 f. 35 g. 40 h. 45 i. 50 j. 55 k. 60 m. 65 n. 70 o. 75

A6. Find the equilibrium downslope speed (m s–1) for the previous problem, if the katabatic air is 5 m thick and the drag coefficient is 0.002.

A7. Find the depth (m) of the thermal internal boundary layer 2 km downwind of the coastline, for an environment with wind speed 8 m s–1 and = 4 K km–1. The surface kinematic heat flux (K·m s–1) is a. 0.04 b. 0.06 c. 0.08 d. 0.1 e. 0.12 f. 0.14 g. 0.16 h. 0.18 i. 0.2 j. 0.22

A8. Assume Tv = 20°C. Find the speed (m s–1) of advance of the sea-breeze front, for a flow depth of 700 m and a temperature excess ∆ (K) of: a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A9. For the previous problem, find the sea-breeze wind speed (m s–1) at the coast.

A10. For a sea-breeze frontal speed of 5 m s–1, find the expected maximum distance (km) of advance of the sea-breeze front for a latitude (°) of a. 10 b. 15 c. 20 d. 25 e. 80 f. 35 g. 40 h. 45 i. 50 j. 55 k. 60 m. 65 n. 70 o. 75

A11. What is the shallow-water wave phase speed (m s–1) for a water depth (m) of: a. 2 b. 4 c. 6 d. 8 e. 10 f. 15 g. 20 h. 25 i. 30 j. 40 k. 50 m. 75 n. 100 o. 200

A12. Assume |g|/Tv = 0.0333 m·s–2·K–1. For a cold layer of air of depth 50 m under warmer air, find the surface (interfacial) wave phase speed (m s–1) for a virtual potential temperature difference (K) of: a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A13. For the previous problem, find the value of the Froude number Fr1. Also, classify this flow as subcritical, critical, or supercritical. Given M = 15 m s–1.


Horizontal divergence (D = ∆U/∆x + ∆V/∆y) is where more air leaves a volume than enters, horizon-tally. This can occur at locations where jet-stream wind speed (Mout) exiting a volume is greater than entrance speeds (Min). Conservation of air mass requires that the num-ber of air molecules in a volume, such as the light blue region sketched in Fig. 13.29, must remain near-ly constant (neglecting compressibility). Namely, volume inflow must balance volume outflow of air. Net vertical inflow can compensate for net hori-zontal outflow. In the troposphere, most of this in-flow happens a mid-levels (P ≈ 50 kPa) as an upward vertical velocity (Wmid). Not much vertical inflow happens across the tropopause because vertical mo-tion in the stratosphere is suppressed by the strong static stability. In the idealized illustration of Fig. 13.29, the inflows [(Min times the area across which the inflow occurs) plus (Wmid times its area of in-flow)] equals the outflow (Mout times the outflow area). The continuity equation describes volume con-servation for this situation as

W D zmid ·∆ (13.15)

or W

Ux

Vy

zmid∆∆

∆∆

·∆ (13.16)

or

WMs

zmid · (13.17)

where the distance between outflow and inflow lo-cations is ∆s, wind speed is M, the thickness of the upper air layer is ∆z, and the ascent speed at 50 kPa (mid tropospheric) is Wmid. Fig. 13.30 shows this scenario for the case-study storm. Geostrophic winds are often nearly parallel to the height contours (solid black curvy lines in Fig. 13.30). Thus, for the region outlined with the black/white box drawn parallel to the contour lines, the main inflow and outflow are at the ends of the box (arrows). The isotachs (shaded) tell us that the in-flow (≈ 20 m s–1) is slower than outflow (≈50 m s–1).

We will focus on two processes that cause hori-zontal divergence of the jet stream: • Rossby waves, a planetary-scale feature for which

the jet stream is approximately geostrophic; and • jet streaks, where jet-stream accelerations cause

non-geostrophic (ageostrophic) motions.

Sample Application Jet-stream inflow winds are 50 m s–1, while out-flow winds are 75 m s–1 a distance of 1000 km further downstream. What updraft is induced below this 5 km thick divergence region? Assume air density is 0.5 kg m–3.

Find the AnswerGiven: Min = 50 m s–1, Mout = 75 m s–1, ∆s = 1000 km, ∆z = 5 km.Find: Wmid = ? m s–1

Use eq. (13.17):Wmid = [Mout - Min]·(∆z/∆s) = [75 - 50 m s–1]·[(5km)/(1000km)] = 0.125 m s–1

Check: Units OK. Physics unreasonable, because the incompressible continuity equation assumes constant density — a bad assumption over a 5 km thick layer.Exposition: Although this seems like a small num-ber, over an hour this updraft velocity would lift air about 450 m. Given enough hours, the rising air might reach its lifting condensation level, thereby creating a cloud or enabling a thunderstorm.

Figure 13.30Over the surface cyclone (X) is a region (box) with faster jet-stream outflow than inflow (arrows). Isotachs are shaded.

20 kPa Heights (m) & Isotachs (m s–1) 12 UTC, 4 Apr 2014

X20

50 m/s

Figure 13.29For air in the top half of the troposphere (shaded light blue), if air leaves faster (Mout) than enters (Min) horizontally, then con-tinuity requires that this upper-level horizontal divergence be balanced by ascent Wmid in the mid troposphere.

Tvp= 15 °C

Find: Ueq= ? m/s

Useeqn.17.9:

where:Δθv=Tvp-Tve=(K)= -5

g= 9.8 m/s^2

Ueq= 17.90 m/s

Check:Unitsok.Physicsok.Discussion:Theequilibriumdownslopewindspeedisthewindspeedatwhichthedragbalancesbuoyancy

A8i)

Given: Tv= 20 °C 293 Kd= 700 mΔθ= 5 K

Find: MSBF= ? m/s

Useeqn.17.11:

wherek= 0.62g= 9.8 m/s^2









b. M = 20 m s–1, ∆T/∆z = –8 °C km–1

c. M = 5 m s–1, ∆T/∆z = –2 °C km–1

d. M = 20 m s–1, ∆T/∆z = 5 °C km–1

e. M = 5 m s–1, ∆T/∆z = –8 °C km–1

f. M = 2 m s–1, ∆T/∆z = –2 °C km–1

g. M = 5 m s–1, ∆T/∆z = 5 °C km–1

h. M = 2 m s–1, ∆T/∆z = –8 °C km–1












However, for most smaller valleys and slopes you can neglect Coriolis force and the V-wind, allowing eq. (17.7) to be solved for some steady-state situa-tions, as shown next. Initially the wind (averaged over the depth of the katabatic flow) is influenced mostly by buoyancy and advection. It accelerates with distance s downslope:

U gT

saveragev

ve· · · sin( )

/1 2 (17.8)

where |g| = 9.8 m·s–2, Tve is absolute temperature in the environment at the height of interest, is the mountain slope angle, and s is distance downslope. The average katabatic wind eventually approach-es an equilibrium where drag balances buoyancy:

U gT

hCeq

v

ve D· · · sin( )

/1 2 (17.9)

where CD is the total drag against both the ground and against the slower air aloft, and h is depth of the katabatic flow.

Katabatic and anabatic winds are part of larger circulations in the valley.

Night At night, the katabatic winds from the bottom of the slopes drain into the valley, where they start to accumulate. This pool of cold air is often stratified like a layer cake, with the coldest air at the bottom and less-cold air on top. Katabatic winds that start higher on the slope often do not travel all the way to the valley bottom (Fig. 17.10). Instead, they either spread out at an altitude where they have the same buoyancy as the stratified pool in the valley, or they end in a turbulent eddy higher above the valley floor. This leads to a relatively mild thermal belt of air at the mid to upper portions of the valley walls — a good place for vineyards and orchards because of fewer frost days. Meanwhile, the cold pool of air in the valley bottom flows along the valley axis in the same di-rection that a stream of water would flow. As this cold air drains out of the valley onto the lowlands, it is known as a mountain wind. This is part of an along-valley circulation. A weak return flow aloft (not drawn), called the anti-mountain wind, flows up-valley, and is the other part of this along-valley circulation.

Sample Application (§) Air adjacent to a 10° slope averages 10°C cooler over its 20 m depth than the surrounding air of virtual temperature 10°C. Find and plot the wind speed vs. downslope distance, and the equilibrium speed. CD = 0.005 .

Find the AnswerGiven: ∆ v= –10°C, Tve=283 K, = 10°, CD= 0.005Find: U (m s–1) vs. x (km), where s ≈ x .

Use eq. (17.9) to find final equilibrium value:

Ueq

( . · )·( )·( )

.·sin(

9 8 10283

200 005

12m s C

Km

001 2

)/

|Ueq| = 15.5 m s–1

Use eq. (17.8) for the initial variation.

Check: Units OK. Physics OK.Exposition: Although these two curves cross, the complete solution to eq. (17.7) smoothly transitions from the initial curve to the final equilibrium value.

Figure 17.10Katabatic winds are cross-valley flows that merge into the along-valley mountain winds draining down valley. Relatively warm air can exist in the “thermal belt” regions.


ecules added to the cool-air column. This creates a pressure gradient near the surface that drives the bottom portion of the sea-breeze circulation. Such hydrostatic thermal circulations were explained in the General Circulation chapter. A sea-breeze front marks the leading edge of the advancing cool marine air and behaves similarly to a weak advancing cold front or a thunderstorm gust front. If the updraft ahead of the front is humid enough, a line of cumulus clouds can form along the front, which can grow into a line of thunderstorms if the atmosphere is convectively unstable. The raised portion of cool air immediately be-hind the front, called the sea-breeze head, is anal-ogous to the head at the leading edge of a gust front. The sea-breeze head is roughly twice as thick as the subsequent portion of the feeder cool onshore flow (which is 0.5 to 1 km thick). The top of the sea-breeze head often curls back in a large horizontal roll eddy over warmer air from aloft. Vertical wind shear at the density interface be-tween the low-level sea breeze and the return flow aloft can create Kelvin-Helmholtz (KH) waves. These breaking waves in the air have wavelength of 0.5 to 1 km. The KH waves increase turbulent drag on the sea breeze by entraining low-momentum air from above the interface. A slowly subsiding return flow occurs over water and completes the circula-tion as the air is again cooled as it blows landward over the cold water. As the cool marine air flows over the land, a thermal internal boundary layer (TIBL) forms just above the ground (Fig. 17.12). The TIBL grows in depth zi with the square root of distance x from the shore as the marine air is modified by the heat flux FH (in kinematic units K·m s–1) from the underlying warm ground:

zFM

xiH2

1 2··

·/

(17.10)

where = ∆ /∆z is the gradient of potential tempera-ture in the air just before reaching the coast, and M is the wind speed. In early morning, the sea-breeze circulation does not extend very far from the coast, but progresses further over land and water as the day progresses. Advancing cold air behind the sea-breeze front be-haves somewhat like a density current or gravity current in which a dense fluid spreads out horizon-tally beneath a less dense fluid. When this is simu-lated in water tanks, the speed MSBF of advance of the sea-breeze front, is

M k g

TdSBF

v

v· ·

∆·

(17.11)

Sample Application What horizontal pressure difference is needed in the bottom part of the sea-breeze circulation to drive a onshore wind that accelerates from 0 to 6 m s–1 in 6 h?

Find the AnswerGiven: ∆M/∆t = (6 m s–1)/(6 h) = 0.000278 m·s–2 Find: ∆P/∆x = ? kPa km–1

Neglect all other terms in the horiz. eq. of motion: ∆M/∆t = –(1/ )·∆P/∆x (10.23a)Assume air density is = 1.225 kg m–3 at sea level. Solve for ∆P/∆x : ∆P/∆x = – · (∆M/∆t) = –(1.225kg m–3)·(0.000278m·s–2) = –0.00034 kg·m–1·s–2/m = –0.00034 Pa m–1

= –0.00034 kPa km–1

Check: Units OK. Magnitude OK.Exposition: Only a small pressure gradient is needed to drive a sea breeze.

Sample Application For a surface kinematic heat flux of 0.2 K m s–1, wind speed of 5 m s–1, and = 3K km–1, find the TIBL depth 5 km from shore.

Find the AnswerGiven: FH=0.2 K·m s–1, M=5 m s–1, =3K km–1, x=5 kmFind: zi = ? m

Use eq. (17.10): zi = [2·(0.2K·m s–1)·(5km) / {(3K km–1)·(5m s–1)} ]1/2 = 0.365 km

Check: Units OK. Magnitude reasonable.Exposition: Above this height, the air still feels the marine influence, and not the warmer land.

Figure 17.13Vertical cross section through a sea-breeze circulation, showing isobars. H and L at any one altitude indicate relatively higher or lower pressure. The horizontal pressure gradient is greatly exaggerated in this illustration. Thick arrows show the coast-normal component of horizontal winds.

MSBF= 6.71 m/s

Check:Unitsok.Physicsok.Discussion:Theseabreezefrontistheboundarybetweentwodifferentairmassescausedbythedifferentspecificheatcapacitiesofwaterandland.Duringtheday,itwilltakelongerforthewatertowarmupcomparingtoland.Thisdifferencecausescoolermarineairtoform,andthetemperaturedifferenceinducesapressuredifferencewhichdrivestheseabreeze.

A12i)

Given: g/Tv= 0.0333 m/s^2/KΔθv= 5 Kh= 50 m

Find: co= ? m/s

Useeqn:17.15:

co= 2.89 m/s

Check:Unitsok.Physicsok.Discussion:Wavesexistintheatmosphere,andbehavesimilarlytowaterwaves.

A13i)

Given: M= 15 m/s

Find: Fr= ?









b. M = 20 m s–1, ∆T/∆z = –8 °C km–1

c. M = 5 m s–1, ∆T/∆z = –2 °C km–1

d. M = 20 m s–1, ∆T/∆z = 5 °C km–1

e. M = 5 m s–1, ∆T/∆z = –8 °C km–1

f. M = 2 m s–1, ∆T/∆z = –2 °C km–1

g. M = 5 m s–1, ∆T/∆z = 5 °C km–1

h. M = 2 m s–1, ∆T/∆z = –8 °C km–1












Waves (vertical oscillations that propagate hori-zontally on the density interface) can exist in air (Fig. 17.16), and behave similarly to water waves. For hy-draulics, if water in a channel is shallow, then long-wavelength waves on the water surface travel at the intrinsic “shallow-water” phase speed co of:

c g ho · (17.14)

where |g| = 9.8 m s–2 is gravitational acceleration magnitude, and h is average water depth. Phase speed is the speed of propagation of any wave crest. Intrinsic means relative to the mean fluid motion. Absolute means relative to the ground. For a two-layer air system, the reduced gravity |g’|= |g|·∆ v/Tv accounts for the cold-air buoyancy relative to the warmer air. For a shallow bottom layer, the resulting intrinsic phase speed of surface waves on the interface between the cold- and warm-air layers is:

c g h gT

hov

v· ·

∆·

/1 2

(17.15)

where ∆ v is the virtual potential temperature jump between the two air layers, Tv is an average absolute virtual temperature (in Kelvin), h is the depth of the cold layer of air, and |g| = 9.8 m·s–2 is gravitational acceleration magnitude. The speed that wave energy travels through a fluid is the group speed cg. Group speed is the speed that hydraulic information can travel relative to the mean flow velocity, and it determines how the upstream flow reacts to downstream flow changes. For a two-layer system with bottom-layer depth less than 1/20 the wavelength, the group speed equals the phase speed cg = co (17.16)

For a statically-stable atmospheric system with constant lapse rate, there is no surface (no interface) on which the waves can ride. Instead, internal waves can exist that propagate both horizontally and vertically inside the statically-stable region. In-ternal waves reflect from solid surfaces such as the ground, and from statically neutral layers. For internal waves, the horizontal component of group velocity ug depends on both vertical and horizontal wavelength . To simplify this compli-cated situation, focus on infinitely-long waves in the horizontal (which propagate the fastest in the horizontal), and focus on a wave for which the verti-cal wavelength is proportional to the depth h of the statically stable layer of air. Thus:

u N hg

T zhg BV

v

v· ·∆∆

·/1 2

(17.17)

Sample Application For a two-layer air system with 5°C virtual poten-tial temperature difference across the interface and a bottom-layer depth of 20 m, find the intrinsic group speed, and compare it to the speed of a water wave.

Find the AnswerGiven: ∆ v = 5°C = 5K, h = 20 m. Assume Tv = 283 KFind: cg = ? m s–1, for air and for water

For a 2-layer air system, use eqs. (17.15 & 17.16): cg = co = [(9.8 m·s–2)·(5K)·(20m)/(283K)]1/2 = 1.86 m s–1

For water under air, use eq. (17.14 & 17.16)): cg = co = [(9.8 m·s–2)·(20m)]1/2 = 14 m s–1

Check: Units OK. Magnitude OK.Exposition: Atmospheric waves travel much slower than channel or ocean waves, and have much longer wavelengths. This is because of the reduced gravity |g’| for air, compared to the full gravity |g| for water.

Figure 17.16Sketch of waves on the interface between cold and warm air lay-ers. is wavelength, and h is average depth of cold air.

Sample Application Find the internal-wave horizontal group speed in air for a constant virtual potential-temperature gradi-ent of 5°C across a stable-layer depth of 20 m.

Find the AnswerGiven: ∆ v = 5°C = 5K, ∆z = h = 20 m. Let Tv = 283 KFind: ug = ? m s–1

Use eq. (17.17): ug = [(9.8 m·s–2)·(5K)/(283K · 20m)]1/2 ·(20m) = 1.86 m s–1

Check: Units OK. Magnitude OK.Exposition: For the example here with ∆z = h, the equation for ug in a stably-stratified fluid is identical to the equation for cg in a two layer system. This is one of the reasons why we can often use hydraulics methods for a stably-stratified atmosphere.









b. M = 20 m s–1, ∆T/∆z = –8 °C km–1

c. M = 5 m s–1, ∆T/∆z = –2 °C km–1

d. M = 20 m s–1, ∆T/∆z = 5 °C km–1

e. M = 5 m s–1, ∆T/∆z = –8 °C km–1

f. M = 2 m s–1, ∆T/∆z = –2 °C km–1

g. M = 5 m s–1, ∆T/∆z = 5 °C km–1

h. M = 2 m s–1, ∆T/∆z = –8 °C km–1











Fromeqn.17.16,forinterfacialwaves:

Alsoforinterfacialwaves,eqn17.18:

Fromthepreviousquestion:

cg=co= 2.89 m/s

Fr= 5.20

Thisflowissupercritical.

Check:Unitsok.Physicsok.Discussion:Airthatissupercriticalismovingsofastthatnoinformationcantravelupwind,sotheupwindairdoesnot'feel'theeffectsofdownwindflowconstrictionsuntilitarrivesattheconstriction.

A17i)

Given: g/Tv= 0.0333 m/s^2/KΔθv= 5 Kh= 300 m

Find: M_gap_max=?m/s

Useeqn.17.25:

Mgap_max= 7.07 m/s









b. M = 20 m s–1, ∆T/∆z = –8 °C km–1

c. M = 5 m s–1, ∆T/∆z = –2 °C km–1

d. M = 20 m s–1, ∆T/∆z = 5 °C km–1

e. M = 5 m s–1, ∆T/∆z = –8 °C km–1

f. M = 2 m s–1, ∆T/∆z = –2 °C km–1

g. M = 5 m s–1, ∆T/∆z = 5 °C km–1

h. M = 2 m s–1, ∆T/∆z = –8 °C km–1












Waves (vertical oscillations that propagate hori-zontally on the density interface) can exist in air (Fig. 17.16), and behave similarly to water waves. For hy-draulics, if water in a channel is shallow, then long-wavelength waves on the water surface travel at the intrinsic “shallow-water” phase speed co of:

c g ho · (17.14)

where |g| = 9.8 m s–2 is gravitational acceleration magnitude, and h is average water depth. Phase speed is the speed of propagation of any wave crest. Intrinsic means relative to the mean fluid motion. Absolute means relative to the ground. For a two-layer air system, the reduced gravity |g’|= |g|·∆ v/Tv accounts for the cold-air buoyancy relative to the warmer air. For a shallow bottom layer, the resulting intrinsic phase speed of surface waves on the interface between the cold- and warm-air layers is:

c g h gT

hov

v· ·

∆·

/1 2

(17.15)

where ∆ v is the virtual potential temperature jump between the two air layers, Tv is an average absolute virtual temperature (in Kelvin), h is the depth of the cold layer of air, and |g| = 9.8 m·s–2 is gravitational acceleration magnitude. The speed that wave energy travels through a fluid is the group speed cg. Group speed is the speed that hydraulic information can travel relative to the mean flow velocity, and it determines how the upstream flow reacts to downstream flow changes. For a two-layer system with bottom-layer depth less than 1/20 the wavelength, the group speed equals the phase speed cg = co (17.16)

For a statically-stable atmospheric system with constant lapse rate, there is no surface (no interface) on which the waves can ride. Instead, internal waves can exist that propagate both horizontally and vertically inside the statically-stable region. In-ternal waves reflect from solid surfaces such as the ground, and from statically neutral layers. For internal waves, the horizontal component of group velocity ug depends on both vertical and horizontal wavelength . To simplify this compli-cated situation, focus on infinitely-long waves in the horizontal (which propagate the fastest in the horizontal), and focus on a wave for which the verti-cal wavelength is proportional to the depth h of the statically stable layer of air. Thus:

u N hg

T zhg BV

v

v· ·∆∆

·/1 2

(17.17)

Sample Application For a two-layer air system with 5°C virtual poten-tial temperature difference across the interface and a bottom-layer depth of 20 m, find the intrinsic group speed, and compare it to the speed of a water wave.

Find the AnswerGiven: ∆ v = 5°C = 5K, h = 20 m. Assume Tv = 283 KFind: cg = ? m s–1, for air and for water

For a 2-layer air system, use eqs. (17.15 & 17.16): cg = co = [(9.8 m·s–2)·(5K)·(20m)/(283K)]1/2 = 1.86 m s–1

For water under air, use eq. (17.14 & 17.16)): cg = co = [(9.8 m·s–2)·(20m)]1/2 = 14 m s–1

Check: Units OK. Magnitude OK.Exposition: Atmospheric waves travel much slower than channel or ocean waves, and have much longer wavelengths. This is because of the reduced gravity |g’| for air, compared to the full gravity |g| for water.

Figure 17.16Sketch of waves on the interface between cold and warm air lay-ers. is wavelength, and h is average depth of cold air.

Sample Application Find the internal-wave horizontal group speed in air for a constant virtual potential-temperature gradi-ent of 5°C across a stable-layer depth of 20 m.

Find the AnswerGiven: ∆ v = 5°C = 5K, ∆z = h = 20 m. Let Tv = 283 KFind: ug = ? m s–1

Use eq. (17.17): ug = [(9.8 m·s–2)·(5K)/(283K · 20m)]1/2 ·(20m) = 1.86 m s–1

Check: Units OK. Magnitude OK.Exposition: For the example here with ∆z = h, the equation for ug in a stably-stratified fluid is identical to the equation for cg in a two layer system. This is one of the reasons why we can often use hydraulics methods for a stably-stratified atmosphere.


where NBV is the Brunt-Väisälä frequency, and ∆ v/∆z is the vertical gradient of virtual potential tem-perature (a measure of static-stability strength).

The ratio of the fluid speed (M) to the wave group velocity (cg , the speed that energy and information travels) is called the Froude number Fr.

Fr = M / cg (17.18)

At least three different Froude numbers can be de-fined, depending on the static stability and the flow situation. We will call these Fr1, Fr2, and Fr3, the last of which will be introduced in a later section. For surface (interfacial) waves in an idealized at-mospheric two-layer system where cg = co = (|g’|·h)1/2, the Froude number is Fr1 = M/co:

FrM

gT

hv

v

1 1 2

·∆

·/ (17.19)

where h is the depth of the bottom (cold) air layer, ∆ v is the virtual potential temperature jump be-tween the two air layers, Tv is an average absolute virtual temperature (in Kelvins), and |g| = 9.8 m·s–2 is gravitational acceleration magnitude. For the other situation of a statically stable region supporting internal waves, the Froude number is Fr2 = M/ug :

FrM

gT z

hv

v2 1 2

·∆∆

·/ (17.20)

The nature of the flow is classified by the value of the Froude number:

• subcritical (tranquil) for Fr < 1 • critical for Fr = 1 • supercritical (rapid) for Fr > 1

For subcritical flow, waves and information travel upstream faster than the fluid is flowing down-stream, thus allowing the upstream flow to “feel” the effect of both upstream conditions and down-stream conditions such as flow constrictions. For supercritical flow, the fluid is moving so fast that no information can travel upstream (relative to a fixed location); hence, the upstream fluid does not “feel” the effects of downstream flow constrictions until it arrives at the constriction. For airflow, the words (upwind, downwind) can be used instead of (up-stream, downstream).

Sample Application For a smooth virtual potential-temperature gradi-ent of 5°C across a stable-layer depth of 20 m, find the Froude number if the average flow speed is 5 m s–1. Discuss whether the flow is critical.

Find the AnswerGiven: ∆ v = 5°C = 5K, ∆z = h = 20 m. Assume Tv = 283 KFind: Fr2 = ? (dimensionless)

Use eq. (17.20). But since ∆z = h , this causes eq. (17.20) to reduce to eq. (17.19). Hence, we get the same answer as in the previous Sample Application: Fr2 = 2.69

Check: Units OK. Magnitude OK.Exposition: This flow is also supercritical. Although this example was contrived to give the same virtual potential temperature gradient across the whole fluid depth as before, often this is not the case. So always use eq. (17.20) for a constant stable stratification flow, and don’t assume that it always reduces to eq. (17.19).

Sample Application For a two-layer air system with 5°C virtual poten-tial temperature difference across the interface and a bottom-layer depth of 20 m, find the Froude number if the average flow speed is 5 m s–1. Discuss whether the flow is critical.

Find the AnswerGiven: ∆ v = 5°C = 5K, h = 20 m. M = 5 m s–1. Assume Tv = 283 KFind: Fr1 = ? (dimensionless)

For a 2-layer air system, use eq. (17.19): Fr1 = (5 m s–1) / [(9.8 m·s–2)·(5K)·(20m)/(283K)]1/2 = (5 m s–1) / (1.86 m s–1) = 2.69

Check: Units OK. Magnitude OK.Exposition: This flow is supercritical. Hence, the air blows the wave downwind faster than it can propagate upwind against the mean flow.


regions due to irregularities in the valley shape, or by obstacles. The resulting turbulence in the hy-draulic jumps causes extra drag, slowing the wind to its critical value. The net result is that many gap winds are likely to have maximum speeds nearly equal to their critical value: the speed that gives Fr = 1. Using this in the definition of the Froude number allows us to solve for the likely maximum gap wind speed through gaps short enough that Coriolis force is not a factor:

M gT

hgap maxv

v·∆

·/1 2

(17.25)

For long gaps, examine the horizontal forces (in-cluding Coriolis force) that act on the air. Consider a situation where the synoptic-scale isobars are nearly parallel to the axis of the mountain range (Fig. 17.22), causing a pressure gradient across the mountains. For long narrow valleys this synoptic-scale cross-mountain pressure gradient is unable to push the cold air through the gap directly from high to low pressure, because Coriolis force tends to turn the wind to the right of the pressure gradient. Instead, the cold air inside the gap shifts its position to en-able the gap wind, as described next. Fig. 17.23 idealizes how this wind forms in the N. Hemisphere. Cold air (light grey in Fig. 17.23a) initially at rest in the gap feels the synoptically im-posed pressure-gradient force FPGs along the valley axis, and starts moving at speed M (shown with the short dark-grey arrow in Fig. 17.23a’) toward the im-posed synoptic-scale low pressure (L) on the oppo-site side of the gap. At this slow speed, both Coriolis force (FCF) and turbulent drag force (FTD) are corre-spondingly small (Fig. 17.23a’). The sum (black-and-white dotted arrow) of all the force vectors (black) causes the wind to turn slightly toward its right in the N. Hemisphere and to accelerate into the gap. This turning causes the cold air to “ride up” on the right side of the valley (relative to the flow direc-tion, see Fig. 17.23b). It piles up higher and higher as the gap wind speed M increases. But cold air is denser than warm. Thus slightly higher pressure (small dark-grey H) is under the deeper cold air, and slightly lower pressure (small dark-grey L) is under the shallower cold air. The result is a cross-valley mesoscale pressure-gradient force FPGm per unit mass m (Fig. 17.23b’) at the valley floor of:

F

mPx

gT

zx

PGm m v

v

1·∆∆

·∆

·∆∆ (17.26)

Figure 17.22Scenario for gap winds in the N. Hemisphere in long valleys, with cold air (light grey) dammed behind a mountain range (dark grey). High pressure (H) is in the cold air, and low pressure (L) is at the opposite side of the mountains. Thick curved lines are sea-level isobars around the synoptic-scale pressure centers. The vertical profile of potential temperature is plotted.

Figure 17.21Cold air flow through a short gap. Vertical slice inside the gap, along section A - A’ from the previous figure. A series of 3 hydraulic jumps are shown in this idealization.

Sample Application Cold winter air of virtual potential temperature –5°C and depth 200 m flows through an irregular short mountain pass. The air above has virtual poten-tial temperature 10°C. Find the max likely wind speed through the short gap.

Find the AnswerGiven: ∆ v = 15°C = 15K, h = 200 m. Find: Mgap max = ? m s–1

For a 2-layer air system, use eq. (17.25): Mgap max = [(9.8 m·s–2)·(15K)·(200m)/(267K)]1/2 = 10.5 m s–1

Check: Units OK. Magnitude OK.Exposition: Because of the very strong temperature jump across the top of the cold layer of air and the cor-respondingly faster wave group speed, faster gap flow speeds are possible. Speeds any faster than this would cause an hydraulic jump, which would increase turbu-lent drag and slow the wind back to this wind speed.

Check:Unitsok.Physicsok.Discussion:Anywindspeedsfasterthanthiswilllikelycauseturbulenceandadragforcethatwillsubsequentlyslowdownthewinds.Also,notethatinthisscenario,weareassumingthatthegapisshortenoughthattheCoriolisforceissmallenoughtobeneglected.

A20a)

Given: g/Tv= 0.0333 m/s^2/KM= 2 m/sΔT/Δz= 5 °C/km 0.005 K/m

Find: λ= ? m

Useeqn.17.30tofindthenaturalwavelength:

Useeqn.5.4atofindtheBrunt-Vasalafrequency:

whereΓd= 9.8 °C/km0.0098 K/m

NBV= 0.0222 /s

λ= 566.05 m

Check:Unitsok.Physicsok.









b. M = 20 m s–1, ∆T/∆z = –8 °C km–1

c. M = 5 m s–1, ∆T/∆z = –2 °C km–1

d. M = 20 m s–1, ∆T/∆z = 5 °C km–1

e. M = 5 m s–1, ∆T/∆z = –8 °C km–1

f. M = 2 m s–1, ∆T/∆z = –2 °C km–1

g. M = 5 m s–1, ∆T/∆z = 5 °C km–1

h. M = 2 m s–1, ∆T/∆z = –8 °C km–1












Figure 17.29Mountain-wave characteristics.

When statically stable air flows with speed M over a hill or ridge, it is set into oscillation at the Brunt-Väisälä frequency, NBV. The natural wave-length is

2 · MNBV

•(17.30)

Longer wavelengths occur in stronger winds, or weaker static stabilities. These waves are known as mountain waves, gravity waves, buoyancy waves, or lee waves. They can cause damaging winds, and interesting clouds (see the Clouds chapter). Friction and turbulence damp the oscillations with time (Fig. 17.29). The resulting path of air is a damped wave:

z zx

bx

12

·exp·

· cos·

(17.31)

where z is the height of the air above its starting equilibrium height, z1 is the initial amplitude of the wave (based on height of the mountain), x is dis-tance downwind of the mountain crest, and b is a damping factor. Wave amplitude reduces to 1/e at a downwind distance of b wavelengths (that is, b· is the e-folding distance).

In the updraft portions of mountain waves, the rising air cools adiabatically. If sufficient moisture is present, clouds can form, called lenticular clouds. The first cloud, which forms over the mountain crest, is usually called a cap cloud (see Clouds chapter). The droplet sizes in these clouds are often quite uniform, because of the common residence times of air in the clouds. This creates interesting optical

Sample Application (§) Find and plot the path of air over a mountain, giv-en: z1 = 500 m, M = 30 m s–1, b = 3, ∆T/∆z = –0.005 K m–1, T = 10°C, and Td = 8°C for the streamline sketched in Fig. 17.29.

Find the AnswerGiven: (see above). Thus T = 283 KFind: NBV = ? s–1, = ? m, and plot z vs. x

From the Stability chapter:

NBV

9 8283

0 005 0 00981 2

. ·· . .

/m s

K

2

= 0.0129 s–1 Use eq. (17.30)

2 300 0129 1

·( ).

m/ss

= 14.62 km

Solve eq. (17.31) on a spreadsheet to get the answer:

Check: Units OK. Physics OK. Sketch OK.Exposition: Glider pilots can soar in the updraft por-tions of the wave, highlighted with white boxes.

Sample Application The influence of the coast mountains extends how far to the west of the coastline in Fig. 17.28? The cold air has virtual potential temperature 8°C colder than the neighboring warm air. The latitude is such that the Coriolis parameter is 10–4 s–1. Mountain height is 2000 m.

Find the AnswerGiven: h = 2000 m, ∆ v = 8°C = 8 K, fc = 10–4 s–1 .Find: R = ? km, external Rossby deformation radius

Assume: |g|/Tv = 0.03333 m·s–2·K–1 The region of influence extends a distance equal to the Rossby deformation radius. Thus, use eq. (17.28): R = [(0.03333 m·s–2·K–1)·(2000 m)·(8 K)]1/2 /(10–4 s–1) = 231,000 m = 231 km

Check: Units OK. Magnitude OK.Exposition: Even before fronts and low-pressure centers hit the coastal mountains, the mountains are already influencing these weather systems hundreds of kilometers offshore.


In summary, a parcel that is displaced vertically from its initial altitude will oscillate up and down. The environment that enables such an oscillation is said to be statically stable. The oscillation is similar to that experienced by a vertically displaced weight suspended between two elastic bands (Fig. 5.13b). The air parcel oscillates vertically with frequency (radians s–1)

Ng

TTzBV

v

vd •(5.4a)

where NBV is the Brunt-Väisälä frequency, and |g| = 9.8 m s–2 is the magnitude of gravitational accel-eration. Virtual temperature Tv must be used to ac-count for water vapor, which has lower density than dry air. If the air is relatively dry, then Tv ≈ T. Use absolute units (K) for Tv in the denominator. The same frequency can be expressed in terms of virtual potential temperature v (see Fig. 5.13c):

Ng

T zBVv

v •(5.4b)

The oscillation period is

PNBV

BV

2 (5.5)

If the static stability weakens (as the environmental ∆T/∆z approaches – d), then the frequency decreases and the period increases toward infinity. The equa-tions above are idealized, and neglect the damping of the oscillation due to air drag (friction).

Create the Governing Equations Consider a scenario as sketched in Fig. 5.13a, where the environment lapse rate is and the dry adiabatic lapse rate is d. Define the origin of a coor-dinate system to be where the two lines cross in that figure, and let z be height above that origin. Consider dry air for simplicity (Tv = T). Based on geometry, Fig. 5.13a shows that the temperature difference between the two lines at height z is ∆T = ( – d)·z. Eq. (5.3b) gives the corresponding vertical force F per unit mass m of the air parcel:

F/m = |g|·∆T/Te = |g|·[( – d)·z]/Te

But F/m = a according to Newton’s 2nd law, where ac-celeration a d2z/dt2. Combining these eqs. gives:

dd

2zt

gT

ze

d2 · ·

Because Te is in Kelvin, it varies by only a small percentage as the air parcel oscillates, so approximate it with the average environmental temperature Te over the vertical span of oscillation. All factors on the right side of the eq. above are constant except for z, so to simplify the notation, use a new variable NBV

2 –|g|·( – d)/ Te to represent those constant factors. [This definition agrees with eq. (5.4a) when you recall that = –∆T/∆z .] This leaves us with a second-order differential equation (a hyperbolic differential eq.) known as the wave equation:

dd

2zt

N zBV22 · (5a)

Solve the Differential Equation For the wave equation, try a wave solution:

z = A·sin( f·t) (5b)

with unknown amplitude A and unknown frequency of oscillation f (radians s–1). Insert (5b) in (5a) to get:

f A f t N A f tBV2 2sin( ) sin( )

After you cancel the A·sin( f·t) from each side, you get: f NBV

Using this in eq. (5b) gives an air parcel height that oscillates in time:

z A N tBV·sin( · ) (5c) Exposition The amplitude A is the initial distance that you displace the air parcel from the origin. If you plug eq. (5.5) into (5c) you get: z = A·sin(2π·t/PBV), which gives one full oscillation during a time period of t = PBV.

Sample Application For a dry standard-atmosphere defined in Chapter 1, find the Brunt-Väisälä frequency & period at z = 4 km.

Find the AnswerGiven: ∆T/∆z = –6.5 K/km from eq. (1.16). Tv = T because dry air. Use Table 1-5 at z = 4 km to get: T = –11°C = 262 KFind: NBV = ? rad s–1, PBV = ? s

Apply eq. (5.4a)

NBV

( . )· – . . ·

9 8262

6 5 9 81

1000

2msK

Kkm

kmmm

= [ 1.234x10–4 s–2]1/2 = 0.0111 rad s–1

Apply eq. (5.5):

PBV2

0 0111radiansradians/s.

= 565.5 s = 9.4 min

Check: Physics and units are reasonable.Exposition: The Higher Math box at right shows that NBV must have units of (radians s–1) because it is the argument of a sine function. But meteorologists often write the units as (s–1), where the radians are implied.

Discussion:oscillationatNBVonlyexistsinastaticallystableenvironment.

A35i)

Tnew= 19 °CTd_new= -1.5 °CPrecip= 0.5 g/kgUseeqn4.14btofindRH:

r@95kPa= 4 g/kgrs@95kPa= 14 g/kgRH@95kPa= 28.57 %

Discussion:Thistypeofprecipitationprocessovermountainsoftenoccursinourregion(thewestcoastofBritishColumbia).Withlow-pressuresystemsthatbringinalotofmoistmarineair,theeastsideoftheVancouverIsland(suchasTofino)oftenreceivesmuchmoreprecipitationthantheLowerMainland.TheairthatmakesittotheLowerMainlandisoftenmuchdrier


A21. For a mountain of width 25 km, find the Froude number Fr3 for the previous problem. Draw a sketch of the type of mountain waves that are likely for this Froude number.

A22. For the previous problem, find the angle of the wave crests, and the wave-drag force per unit mass. Assume H = 1000 m and hw = 11 km.

A23(§). Plot the wavy path of air as it flows past a mountain, given an initial vertical displacement of 300 m, a wavelength of 12.5 km, and a damping fac-tor of a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A24(§) Given a temperature dew-point spread of 1.5°C at the initial (before-lifting) height of air in the previous problem, identify which wave crests con-tain lenticular clouds.

A25. Cold air flow speed 12 m s–1 changes to 3 m s–1 after a hydraulic jump. Assume |g|/Tv = 0.0333 m·s–2·K–1. How high can the hydraulic jump rise if the exit velocity (m s–1) is a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A26. Assuming standard sea-level density and streamlines that are horizontal, find the pressure change given the following velocity (m s–1) change: a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A27. Wind at constant altitude decelerates from 12 m s–1 to the speed (m s–1) given below, while passing through a wind turbine. What opposing net pres-sure difference (Pa) would have caused the same de-celeration in laminar flow? a. 1.0 b. 1.5 c. 2.0 d. 2.5 e. 3.0 f. 3.5 g. 4.0 h. 4.5 i. 5.0 j. 5.5 k. 6.0 m. 7 n. 8 o. 9

A28. Air with pressure 100 kPa is initially at rest. It is accelerated isothermally over a flat 0°C snow surface as it is sucked toward a household venti-lation system. What is the final air pressure (kPa) just before entering the fan if the final speed (m s–1) through the fan is: a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 j. 10 k. 11 m. 12 n. 13 o. 14 p. 15

A29. A short distance behind the jet engine of an aircraft flying in level flight, the exhaust tempera-ture is 500°C. After the jet exhaust decelerates to zero, what is the final exhaust air temperature (°C),

neglecting conduction and mixing, assuming the initial jet-blast speed (m s–1) is: a. 100 b. 125 c. 150 d. 175 e. 200 f. 210 g. 220 h. 230 i. 240 j. 250 k. 260 m. 270 n. 280 o. 290

A30. An 85 kW electric wind machine with a 3 m radius fan blade is used in an orchard to mix air so as to reduce frost damage on fruit. The fan horizon-tally accelerates the air from calm to the speed (m s–1) given below. Find the temperature change (°C) across the fan, neglecting mixing with the environ-mental air. a. 6 b. 6.5 c. 7 d. 7.5 e. 8 f. 8.5 g. 9 h. 9.5 i. 10 j. 10.5 k. 11 m. 12 n. 13 o. 14

A31. Tornadic air of temperature 25°C blows with speed (m s–1) given below, except that it stagnates upon hitting a barn. Find the final stagnation tem-perature (°C) and pressure change (kPa). a. 100 b. 125 c. 150 d. 175 e. 200 f. 210 g. 220 h. 230 i. 240 j. 250 k. 260 m. 270 n. 280 o. 290

A32. Find the speed of sound (m s–1) and Mach number for Mair = 100 m s–1, given air of tempera-ture (°C): a. –50 b. –45 c. –40 d. –35 e. –30 f. –25 g. –20 h. –15 i. –10 j. –5 k. 0 m. 5 n. 10 o. 15 p. 20

A33. Water flowing through a pipe with speed 2 m s–1 and pressure 100kPa accelerates to the speed (m s–1) given below when it flows through a constric-tion. What is the fluid pressure (kPa) in the constric-tion? Neglect drag against the pipe walls. a. 6 b. 6.5 c. 7 d. 7.5 e. 8 f. 8.5 g. 9 h. 9.5 i. 10 j. 10.5 k. 11 m. 12 n. 13 o. 14

A34. For the bora Sample Application, redo the cal-culation assuming that the initial inversion height (km) is: a. 1.1 b. 1.15 c. 1.25 d. 1.3 e . 1.35 f. 1.4 g. 1.45 h. 1.5 i. 1.55 j. 1.6 k. 1.65 m. 1.7 n. 1.75 o. 1.8 p. 1.85

A35. Use a thermodynamic diagram. Air of ini-tial temperature 10°C and dew point 0°C starts at a height where the pressure (kPa) is given below. This air rises to height 70 kPa as it flows over a mountain, during which all liquid and solid water precipitate out. Air descends on the lee side of the mountain to an altitude of 95 kPa. What is the temperature, dew point, and relative humidity of the air at its fi-nal altitude? How much precipitation occurred on the mountain? [Hint: use a thermo diagram.] a. 104 b. 102 c. 100 d. 98 e. 96 f. 94 g. 92 h. 90 i. 88 j. 86 k. 84 m. 82 n. 80

92 CHAPTER 4 • WATER VAPOR

Sample Application Find Td and RH% for r= 10 g kg–1, P= 80 kPa, T=20°C.

Find the AnswerGiven: r = 0.01 gvapor gair

–1, P = 80 kPa, T = 20°CFind: Td = ? °C, and RH% = ? %

Use eq. (4.15b): Td = [(1/(273.15K) – (1.844x10–4 K–1)· ln{ [(80kPa)·(0.01gvapor gair

–1)] / [(0.6113kPa)·((0.01g g–1) +(0.622g g–1)] }]–1 =[(1/(273.15K)–(1.844x10–4K–1)·ln(2.07)]–1 = 283.78K. Thus, Td = 283.78 – 273.15 = or 10.6°C

At Td = 10.6°C, eq. 4.1b gives e ≈ 1.286 kPa At T = 20°C, Table 4-1 gives es = 2.369 kPa Use eq. (4.14a): RH% = 100%·(1.286 kPa/2.369 kPa) RH% = 54.3%

Check: Magnitude and units are reasonable.

Table 4-2b. Moisture variables. (continuation)

Variable name: Relative Humidity Dewpoint

Symbol: RH or RH% Td

Units: (dimensionless) (K)Alternative Units: (%) (°C)

Defining Equation:(& equation number) RH

ees

or RH ees

%%100

(4.14a)

“Temperature to which a given air parcel must be cooled at constant pressure and constant water-vapor content in order for saturation to occur.”*

Alternative Definitions:

RHqq

rrs

v

vs s (4.14b)

RH qq

rrs

v

vs s

%%100

(4.14c)

T

T Leed

o

v

o

11

· ln (4.15a)

T

T Lr P

e rdo

v

o

11

ln·

·( ) (4.15b)

If Saturated: RH = 1.0 or RH% = 100% (4.14d) Td = T

Key Constants: eo = 0.6113 kPa, To = 273.15 Kv/Lv =1.844x10–4 K–1.

= d/ v = 0.622 gvapor gdry air–1

Typical Values: RH = 0.0 to 1.0 or RH% = 0% to 100% See Table 4-1. Td ≤ T Relevance: • regulates the max possible evaporation

into the air.• easy to measure via: (1) capacitance changes across a plastic dielectric; (2) electrical resistance of an emulsion made of carbon powder; or (3) organic-fiber con-traction/expansion.• most used by the general public.

• easy to measure via cooling a mirror to the point where condensation (dew) forms on it. The temperature at which this first happens is the dewpoint, as detected by measuring how well a light beam can re-flect off the mirror.• a very accurate method for humidity measurement.

Notes: • it is possible to have relative humidities as high as about 100.5%. This is called supersaturation (see the Precipitation Processes chapter).

• Td also called dewpoint temperature.• (T - Td) = dewpoint depression = temperature dewpoint spread* Glickman, T. S., 2000: Glossary of Meteorol-ogy. American Meteorological Society.

Figure 4.3Relationship between relative humidity and mixing ratio at P = 101.325 kPa. Grey region is unsaturated. White region is unphysical (except sometimes for 0.5% of supersaturation).

444 QQMZ - University of British Columbia · following radius of curvature (km). Assume the air...

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