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UHM Course ATMO402 - Ch3 - Gradient Wind Approximation & Thermal Wind

Learn advanced concepts in dynamics include vorticity, cyclogenesis, jet streams, fronts, mesoscale circulations and so on.

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UHM Course ATMO402 - Ch3 - Gradient Wind Approximation & Thermal Wind

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  1. ATMO402 “Applied Atmospheric Dynamics” Chapter 3 Elementary Applications of the Basic Equations • Pressure as a Vertical Coordinate • Basic Equations in Isobaric Coordinates • Natural Coordinates • Balanced Flow Geostrophic Flow Inertial Flow Cyclostrophic Flow The Gradient Wind Approximation • The Thermal Wind Sen Zhao Email:zhaos@hawaii.edu

  2. Horizontal momentum equation in Nature Coordinates Acceleration Forces Acceleration Tangent Centripetal Coriolis force Coriolis force ⊥ t Pressure gradient force 2

  3. Balanced Forces 0 = −%& '− (% −)Φ , )+ Centrifugal force • An apparent force • Negative sign means it is always directed radially outwards from that axis −%& ', Coriolis force • In the NH (f>0), directed to the right of the wind direction • In the SH (f<0), directed to the left of the wind direction −(%, Pressure gradient force −)- )+, • Directed from high pressure (geopotential) to low pressure (geopotential) 3

  4. Cyclone/Anticyclone vs. Radius of curvature Southern Hemisphere (. < 0) Northern Hemisphere (. > 0) 2 2 , , Counterclockwise Anti-cyclonic flow Cyclonic flow , , Clockwise 2 2 Anti-cyclonic flow Cyclonic flow 4

  5. 0 = −%& '− (% −)Φ Inertial flow , )+ Southern Hemisphere (. < 0) Northern Hemisphere (. > 0) 2 2 X Coriolis , , 3 > 0 Centrifugal Coriolis Centrifugal Counterclockwise Anti-cyclonic flow Inertial flow can be only anti-cyclonic Cyclonic flow X Coriolis 3 < 0 Clockwise , , Coriolis Centrifugal Centrifugal 2 2 Anti-cyclonic flow Cyclonic flow 5

  6. 0 = −%& '− (% −)Φ Cyclostrophic flow , )+ )Φ )+> 0 2 )Φ )+< 0 2 X PGF , H , L Centrifugal 3 > 0 Centrifugal PGF Counterclockwise Cyclonic flow Cyclonic flow )Φ )+< 0 )Φ )+> 0 X H PGF L 3 < 0 Clockwise , , Centrifugal Centrifugal PGF 2 Anti-cyclonic flow 2 Anti-cyclonic flow 6

  7. Gradient wind A three-way balance among the Coriolis, centrifugal, and pressure gradient forces. 0 = −%& '− (% −)Φ , )+ 1 '%&+ (% +)Φ Rewrite as )+= 0 67&+ 87 + 9 = 0 Looks like quadratic Roots of the quadratic formula 7

  8. 1 '%&+ (% +)Φ 6 =1 Solution for gradient wind )+= 0 ',8 = (,9 =)Φ Substituting with )+ The solutions in nature coordinates: =/& (&'& 4 % = −(' − ')Φ 2± )+

  9. 0 = −%& '− (% −)Φ Gradient Wind Approximation , )+ Northern Hemisphere (. > 0) )Φ )+> 0 2 2 )Φ )+< 0 X Coriolis , H PGF Coriolis , L Centrifugal 3 > 0 PGF Centrifugal Counterclockwise Cyclonic high Cyclonic low (regular) )Φ )+< 0 )Φ )+> 0 H PGF L 3 < 0 Clockwise Coriolis , Coriolis , Centrifugal Centrifugal PGF 2 2 Anti-cyclonic low (anomalous) Anti-cyclonic high 9

  10. Cyclonic low in the NH Northern Hemisphere (. > 0) 500 mb @A @B< 0, ' > 0 2 Coriolis , L PGF Centrifugal Cyclonic low (regular) L =/& (&'& 4 % = −(' − ')Φ 2+ )+ 10

  11. Baric Flow Baric flow: the Coriolis and pressure gradient forces are oppositely directed. However, anomalous low is anti-baric flow. )Φ )+> 0 Recalling the geostrophic approximation 0 = −(%C−)Φ 3 < 0 , )+ L Anti-cyclonic low (anomalous) X Coriolis the geostrophic wind Vgis negative and is clearly not a useful approximation to the actual speed , Centrifugal PGF 2 Northern Hemisphere (. > 0) 11

  12. 0 = −%& '− (% −)Φ Anticyclonic high , )+ @A @B< 0, ' < 0 Northern Hemisphere (. > 0) =/& (&'& 4 % = −(' − ')Φ =/& (&'& 4 % = −(' − ')Φ 2+ 2− )+ )+ H PGF H PGF , , Coriolis Centrifugal Coriolis Centrifugal 2 2 Anomalous High Regular High For both regular and anomalous highs, it is required (&'& 4 <(&' )- )+ − ')Φ )+> 0 4 As |R| à 0, pressure gradient approach zero. The pressure field near the center of a high is always flat, the wind is usually gentle. 12

  13. Anticyclonic high 13

  14. Why we call “regular” or “anomalous” ? The absolute angular momentum about the axis of rotation for the circularly symmetric motions is given by VR + f R2/2. NH SH absolute angular momentum > 0 regular anomalous absolute angular momentum < 0 anomalous regular • Regular gradient wind balances have positive absolute angular momentum in the NH. • Anomalous cases have negative absolute angular momentum 14

  15. Absolute angular momentum of gradient wind =/& (&'& 4 % = −(' − ')Φ Gradient wind 2± )+ =/& (&'& 4 − ')Φ %' = −('&/2 ± ' Times R both sides )+ =/& %' +('& (&'& 4 − ')Φ Absolute angular momentum of gradient wind = ±' 2 )+ NH SH Regular: absolute angular momentum > 0 anomalous R>0, select + root (C) R<0, select – root (A) Regular: absolute angular momentum < 0 anomalous R>0, select – root (A) R<0, select + root (C) 15

  16. Gradient wind vs. Geostrophic wind (1) • Directions are the same, is parallel to the isobars, low pressure in its left in NH. −%& '− (% −)Φ (%C= −)Φ )+= 0 %C %= 1 + % ('= 1 + 'D −%& '− (% + (%C= 0 )+ • For normal cyclonic flow ( f R > 0), Vg> V , whereas for anticyclonic flow ( f R < 0), Vg< V . • Therefore, the geostrophic wind is an overestimate in a region of cyclonic curvature and an underestimate in a region of anticyclonic curvature. 16

  17. Gradient wind vs. Geostrophic wind (2) %C %= 1 + % ('= 1 + 'D the magnitude of V /( f R ) is just the Rossby number. Small Ro • For mid-latitude synoptic systems, Ro ~ 0.1, the difference between gradient and geostrophic wind speeds generally < 10 to 20%. Large Ro • f is small region, for the tropical disturbances, the Ro ~ 1-10, and the gradient wind formula must be applied rather than the geostrophic wind. • R is small, such as tornadoes, Ro ~ 103, cyclostrophic flow is dominated. • Highly curved wind. 17

  18. The thermal wind Holton’s Text Book “The geostrophic wind must have vertical shear in the presence of a horizontal temperature gradient, as can be shown easily from simple physical considerations based on hydrostatic equilibrium.” 18

  19. A illustration of thermal wind • The thickness between the pressure surface is larger at higher temperature place than that at lower temperature place Thickness between two isobars is proportional to the mean temperature in the layer. • The horizontal gradient of geopotential height becomes larger in upper level than over lower level )Φ )7 F+GH6IHI JFKℎ ℎHFMℎK • Geostrophic wind must be greater over the upper level than that of lower level. Tx1 < Tx2 NC=1 )Φ )7 (

  20. The geostrophic wind must have vertical shear in the presence of a horizontal temperature gradient is saying that Change in the geostrophic wind with respect to )OC )P height, i.e., QRS depends on horizontal temperature gradient How? 20

  21. Deriving the thermal wind relationship The thermal wind relationship can be derived by Differentiating with respect to pressure: The geostrophic approximation TC= −1 ( OC=1 (V×QRΦ )NC )P=1 )TC )P= −1 ) )7 )Φ )P ) )U )Φ )P )Φ )U ,NC=1 )Φ )7 ( ( ( )NC )P=1 )TC )P= −1 ) )7 −'S ) )U −'S ( P ( P P)TC P)NC )P=' )S )U )P= −' )S )7 The hydrostatic approximation ( ( )Φ )P= −X = −'S The thermal wind equation )lnP=' ( )U P )NC )lnP= −' )TC )S )7 )S ( )OC )lnP= −' (V×QRS

  22. The thermal wind vector O[ A relationship for the vertical wind shear (i.e., the rate of change of the geostrophic wind with respect to ln p). • )OC )lnP= −' (V×QRS The vector difference between geostrophic winds at two levels. • Integrating thermal wind equation from pressure level p0to level p1to get R` O[≡ OCP= − OCP] = −' (^ V×QRS_ lnP Ra Letting S denote the mean temperature in the layer between pressure level p0 and level p1. O[= −' (V×QRS lnP] +bKH −lnP= = lnP] > 0 P] P= P=

  23. The thermal wind vector O[ T[= −' ) S )UlnP] O[= −' (V×QRS lnP] ( P= P= N[=' ) S )7lnP] ( P= S − cS Cold O[parallel to the isotherms with the cold air at the left and warm air at the right of the thermal wind direction in NH −QRS S O[ Warm 23

  24. The thermal wind vector O[ Finding <T> itself is kind of a mess But thickness is not a problem at all! The thickness of the layer from P]to P=is just Φ=− Φ] So, the thermal wind in terms of horizontal gradient of thickness T[= −1 ) )UΦ=− Φ] O[≡ OCP= − OCP] = ( = dV×QRΦ=− Φ] N[=1 ) )7 Φ=− Φ] (

  25. Thickness vs. Mean temperature The hydrostatic approximation )Φ )P= −X = −'S P Φ=− Φ]= Me[= −' S lnR` Ra = ' S lnRa R` The thickness is proportional to the mean temperature in the layer. Hence, lines of equal e[(isolines of thickness) are equivalent to the isotherms of mean temperature in the layer. 25

  26. Thermal wind and temperature advection Cold advection, backing with height (wind turns counterclockwise) Geostrophic wind on Level 1 Geostrophic wind on Level 0 Thermal wind: Change of geostrophic wind between Level 0 and 1 26

  27. Warm advection, veering with height (wind turns clockwise) Thermal wind: Change of geostrophic wind between Level 0 and 1 Geostrophic wind on Level 1 Geostrophic wind on Level 0 27

  28. 28

  29. Vg @ 1000 hPa Vg @ 500 hPa VT 29

  30. Thermal Wind • Thermal wind blows parallel to the isotherms (thickness contours) with the cold air at the left and warm air at the right of the wind direction in NH. • Thermal wind strength is proportional to mean temperature gradient. • A geostrophic wind turning counterclockwise with height is associated with cold-air advection. • Clockwise turning (veering) of the geostrophic wind with height implies warm advection. Implications • The horizontal temperature advection can be estimated from the sounding of wind at a given location. • The geostrophic wind at any level can be calculated from the mean temperature.

  31. 31

  32. Temperature Gradient and Geostrophic Wind • Text Book Example • For example, if the geostrophic wind at 850 hPa is known and the mean horizontal temperature gradient in the layer 850 to 500 hPa is also known, the thermal wind equation can be applied to obtain the geostrophic wind at 500 hPa. OC500 ℎh6 = OC850 ℎh6 + O[ = OC850 ℎh6 −' (V×QRS ln850 500

  33. horizontal temperature advection? 33

  34. 34

  35. Where is jet core? T[= −' ) S )UlnP] ( P= Zonal mean temperature December-February 1 Kelvin 305 - - 260 270 245 ) S )U + − 260 245 300 2 295 230 3 260 245 290 4 285 5 245 6 230 280 - 8 275 230 10 + 210 T[ 270 245 Pressure (hPa) 230 265 20 260 30 255 40 210 250 210 50 + - 245 60 − − 240 80 230 100 235 210 230 210 210 225 200 230 220 230 245 + 300 245 210 230 260 - 400 230 260 200 + + 245 260 500 270 270 245 600 190 285 800 260 180 285 270 60 1000 80 ON 60 ON 40 ON 20 ON 0 20 OS 40 OS OS 80 OS O y direction f<0 f>0

  36. Where is jet core? T[= −' ) S )UlnP] ( P= Zonal mean wind December-February − 1 m/sec 100 - - -15 -30 -50 ) S )U + 15 30 -2 15 80 2 60 0 3 50 4 40 -30 -2 5 35 2 6 30 - 8 25 10 + 30 20 -15 T[ -15 15 Pressure (hPa) 10 0 20 5 15 -2 30 2 -2 15 0 -2 40 0 50 -5 + - 60 2 2 − − -10 -2 80 -15 15 2 100 2 -20 15 0 -25 -30 200 -35 300 -40 0 15 - -50 400 40 + 15 + + 500 2 -60 600 -80 2 0 800 -2 2 -100 0 2 1000 80 ON 60 ON 40 ON 20 ON 0 20 OS OS 60 OS 80 OS O y direction f<0 f>0

  37. Ocean surface currents 37

  38. Given a geopotential height field, how to estimate the geostrophic wind?

  39. In p coordinate, for p=1000 hPa δφ = −300*10⋅m2/ s2 δx = 750km = 7.5*10 f = 2*7.29*10−5*sin(ϕ) =1.0*10−4 vg=∂φ fδx= −40m / s 5m f∂x=δφ Vg > actual wind δφ =100*10⋅m2/ s2 6m δy =1500km =1.5*10 f = 2*7.29*10−5*sin(ϕ) = 0.6*10−4 ug= −∂φ f∂y= −δφ ~ actual wind fδy= −11m / s

  40. 0 = −%& '− (% −)Φ Gradient Wind in the SH , )+ Southern Hemisphere (. < 0) )Φ )+> 0 2 2 )Φ )+< 0 , H , L 3 > 0 Counterclockwise )Φ )+< 0 )Φ )+> 0 H L 3 < 0 Clockwise , , 2 2 40

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