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AOSS 401 Geophysical Fluid Dynamics: Atmospheric Dynamics Prepared: 20131001 Balanced Flows

AOSS 401 Geophysical Fluid Dynamics: Atmospheric Dynamics Prepared: 20131001 Balanced Flows. Richard B. Rood (Room 2525, SRB) rbrood@umich.edu 734-647-3530 Cell: 301-526-8572. Class News. Ctools site ( AOSS 401 001 F13 ) First Examination on October 22, 2013

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AOSS 401 Geophysical Fluid Dynamics: Atmospheric Dynamics Prepared: 20131001 Balanced Flows

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  1. AOSS 401Geophysical Fluid Dynamics:Atmospheric DynamicsPrepared: 20131001Balanced Flows Richard B. Rood (Room 2525, SRB) rbrood@umich.edu 734-647-3530 Cell: 301-526-8572

  2. Class News • Ctools site (AOSS 401 001 F13) • First Examination on October 22, 2013 • Second Examination on December 10, 2013 • Homework posted: • Ctools Assignments tab • Due Thursday October 10, 2013 • Derivations (using notes)

  3. Weather • National Weather Service • Model forecasts: • Weather Underground • Model forecasts: • NCAR Research Applications Program

  4. Outline • Geostrophic Balance and the Real Wind • Natural Coordinates • Balanced flows • Geostrophic • Cyclostrophic • Gradient

  5. Atmosphere in Balance • Hydrostatic balance (no vertical acceleration) • Geostrophic balance (no horizontal acceleration or divergence) • Adiabatic lapse rate (no clouds or precipitation) • Vertical motion is weak for large-scale flow. • A good approximation is that flow horizontal. • In the upper atmosphere the flow is often adiabatic.

  6. Scale Analysis • Scale analysis is reliant on observations of the preferred motion of the fluid. • What are the size, spatial scale, of the motions? • What are the time scales of the motions? • How did we define large? For our flow we compared f(rotation) to U/L

  7. I present these equations: Assume no viscosity and no vertical motion Pressure What is vertical coordinate? What conservation law? Momentum

  8. Geostrophic balance Low Pressure High Pressure Flow initiated by pressure gradient Flow turned by Coriolis force

  9. Geostrophic & observed wind 300 mb

  10. Describe previous figure. What do we see? • At upper levels (where friction is negligible) the observed wind is parallel to geopotential height contours. • (On a constant pressure surface) • Wind is faster when height contours are close together. • Wind is slower when height contours are farther apart. • There is curvature in the flow • Hence acceleration

  11. Consider this simple “map” • Contours of geopotential • Upper troposphere • Note coordinate system • What is the direction of the geostrophic wind? • Where is it weaker or stronger?

  12. Geopotential (Φ) in upper troposphere ΔΦ > 0 north Φ0+ΔΦ Φ0 Φ0+2ΔΦ Φ0+3ΔΦ east south west

  13. Geopotential (Φ) in upper troposphere ΔΦ > 0 north Φ0+ΔΦ Φ0 Δy Φ0+2ΔΦ Φ0+3ΔΦ east south west

  14. Geopotential (Φ) in upper troposphere ΔΦ > 0 north Φ0+ΔΦ δΦ = Φ0 – (Φ0+2ΔΦ) Φ0 Δy Φ0+2ΔΦ Φ0+3ΔΦ east south west

  15. Geopotential (Φ) in upper troposphere ΔΦ > 0 north Φ0+ΔΦ Φ0 Δy Φ0+2ΔΦ Φ0+3ΔΦ east south west

  16. The horizontal momentum equation Assume no viscosity

  17. Geostrophic approximation

  18. Geopotential (Φ) in upper troposphere ΔΦ > 0 north Φ0+ΔΦ Φ0 Δy Φ0+2ΔΦ Φ0+3ΔΦ east south west

  19. Geopotential (Φ) in upper troposphere ΔΦ > 0 north Φ0+ΔΦ Φ0 Δy Φ0+2ΔΦ Φ0+3ΔΦ east south west

  20. Think about this for a minute …Geopotential (Φ) in upper troposphere ΔΦ > 0 T gradient north Φ0+3ΔΦ W cf Φ0+2ΔΦ Δy Φ0+ΔΦ pgf Φ0 C east south west SOUTHERN HEMISPHERE

  21. How did we get the wind? • This is the i (east-west, x) component of the geostrophic wind. • We have estimated the derivatives based on finite differences. • Does this seem like a reverse engineering of the methods we used to derive the equations? • There is a consistency • The direction comes out correctly! (towards east) • The strength is proportional to the gradient.

  22. The geostrophic wind can only be equal to the real wind if the height contours are straight. north Φ0+ΔΦ Φ0 Φ0+2ΔΦ Φ0+3ΔΦ east south west

  23. Geostrophic & observed wind 300 mb

  24. Geopotential (Φ) in upper troposphere • Think about the observed wind • Flow is parallel to geopotential height lines • There is curvature in the flow Where is the effect of curvature here?

  25. Geopotential (Φ) in upper troposphere • Think about the observed (upper level) wind • Flow is parallel to geopotential height lines • There is curvature in the flow • Geostrophic balance describes flow parallel to geopotential height lines • Geostrophic balance does not account for curvature • What’s one way to think about this? • Curvature means there is acceleration • How to best describe balanced flow with curvature?

  26. Another Coordinate System? • We want to simplify the equations of motion • For horizontal motions on many scales, the atmosphere is in balance • Mass (p, Φ) fields in balance with wind (u) • It is easy to observe the pressure or geopotential height, much more difficult to observe the wind • Need to describe balance between pressure gradient, Coriolis and curvature

  27. “Natural” Coordinate System • Follow the flow • From hydrodynamics—assumes no local changes (“steady state”) • No local change in geopotential height • No local change in wind speed or direction • Assume • Horizontal flow only (no vertical component) • No friction • This is like a Lagrangian parcel approach

  28. Return to Geopotential (Φ) in upper troposphere Define one component of the horizontal wind as tangent to the direction of the wind. t north Φ0 t t t Φ0+3ΔΦ east south west ΔΦ > 0

  29. How do these natural coordinates relate to the tangential coordinates? • They are still tangential, but the unit vectors do not point west to east and south to north. • The coordinate system turns with the wind. • And if it turns with the wind, what do we expect to happen to the forces? Ω a Φ = latitude Earth

  30. Looking down from above

  31. Looking down from above

  32. Looking down from above

  33. Looking down from above

  34. Looking down from above

  35. Return to Geopotential (Φ) in upper troposphere Define the other component of the horizontal wind as normal to the direction of the wind. n north Φ0 n n t n t t Φ0+3ΔΦ east south west ΔΦ > 0

  36. “Natural” Coordinate System • Regardless of position (i,j) • t always points in the direction of flow • n always points perpendicular to the direction of the flow toward the left • Remember the “right hand rule” for vectors? Take k x t to get n • Assume • Pressure as a vertical coordinate • Flow parallel to contours of geopotential height

  37. “Natural” Coordinate System • Advantage: We can look at a height (on a pressure surface) and pressure (on a height surface) and estimate the wind. • It is difficult to directly measure winds • We estimate winds from pressure (or hydrostatically equivalent height), a thermodynamic variable. • Natural coordinates are useful for diagnostics and interpretation.

  38. “Natural” Coordinate System • For diagnostics and interpretation of flows, we need an equation…

  39. Return to Geopotential (Φ) in upper troposphere north Low n n t n t t • Geostrophic assumption. • Do you notice that those n vectors point towards something out in the distance? HIGH east south west ΔΦ > 0

  40. Return to Geopotential (Φ) in upper troposphere Do you see some notion of a radius of curvature? Sort of like a circle, but NOT a circle. north Low n n t n t HIGH t east south west

  41. Time to look at themathematics One direction: no (u,v) components First simplification: the velocity • Always positive • Always points in the positive t direction Define velocity as: Definition of magnitude:

  42. Goal: Quantify Acceleration acceleration is: (Product Rule) Change in speed Change in Direction

  43. How to get as a function of V, R ?

  44. Remember our circle geometry… Δs=RΔφ this is not rotation of the Earth! It is an element of curvature in the flow. Δφ Δt t+Δt t R= radius of curvature Δs t

  45. Remember our circle geometry… Δs=RΔφ this is not rotation of the Earth! It is an element of curvature in the flow. Δφ Δt t+Δt n t R= radius of curvature n Δs t

  46. Remember our circle geometry… Δs=RΔφ If Δs is very small, Δt is parallel to n. So, Δt points in the direction of nas Δs  0 Δφ Δt t+Δt n t R= radius of curvature n Δs t

  47. Remember, we want an expression for From circle geometry we have: Rearrange and take the limit Use the chain rule Remember the definition of velocity

  48. Goal: Quantify Acceleration acceleration defined as: (Product Rule) We just derived: So the total acceleration is

  49. Acceleration in Natural Coordinates Along-flow speed change ?

  50. Acceleration in Natural Coordinates The total acceleration is Definition of wind speed angle of rotation Circle geometry angular velocity Plug in for Δs Centrifugal force

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