1. GEU 0047: Meteorology Lecture 8 Air Pressure and Winds

2. How can you know this wind direction?

3. Pressure The steady drumbeat of countless numbers of atoms and molecules, exchanging momentum with the walls of a container and providing the “Pressure”.

4. Gas Law The pressure is related to the density and temperature of the gas through its internal energy. –Kinetic Energy = 1/2 m v 2 –Exchange of Momentum, the drumbeat of atoms and molecules. Equation of State –PV = NkT N = number k = Boltzman's constant = 1.38 x 10 -23 J/K –PV = nRT n = grams per mole / molar mass R = Gas constant = 8.31 Joules/mole/K

5. Partial Pressures Pressure = P N2 + P O2 + + P Ar + P H20 + P CO2 +... Pressure = Force / Area (Snow shoes on thin ice, high heels on asphalt) Force = Weight of overlying air = mass x gravity

6. Weight Pressure = Force / Area Force = Weight of overlying column of air = mass x gravity

7. Pressure Units Units of Pressure Force / Area = Newton/m 2 = Pascal Weather RelatedSea Level5,000 feet10,000 feet Pascals101,325 Millibars1013.25500250 Atmospheres1.00.50.2 Inches of Mercury  30.015.07.5 PSI14.77.43.0

8. Pressure Records

10. Fluid Pressure at Height (or Depth) Since pressure (P) = Force / Area and the force is the weight (W = mg) of the overlying column... P = mg/A Multiply the top and bottom by the volume (V) P = Vmg/VA m/V is the density and V/A is just the height for a Therefore, P =  g h

11. Pressure and Columns At what depth do we encounter a pressure = 1 atm. In a) water, b) gasoline, c) mercury?

12. Pressure and Columns At what depth do we encounter a pressure = 1 atm. In a) water, b) gasoline, c) mercury? 1 atm = 101325 Pascals P =  ghh a = P/  a g h b = P/  b g h c = P/  c g

13. Pressure and Columns At what depth do we encounter a pressure = 1 atm. In a) water, b) gasoline, c) mercury? 1 atm = 101325 Pascals P =  ghh a = P/  a g= 101325/1000*9.8 h b = P/  b g = 101325/680*9.8 h c = P/  c g = 101325/13600*9.8

14. Inches of Mercury At what depth do we encounter a pressure = 1 atm. In a) water, b) gasoline, c) mercury? 1 atm = 101325 Pascals P =  ghh a = P/  a g= 10.3 meters h b = P/  b g = 15.2 meters h c = P/  c g = 0.76 meters 1 atm ~ 30 inches mercury (0.76 meters)

15. Pressure and Altitude There is a large difference in pressure with altitude.

16. Pressure Changes Horizontal: Changes ~ 1 mb over 10,000s of meters Vertical: 1 mb over 10s of meters Barometric Pressure must be corrected for Altitude (and Temperature). Vertical atmospheric motions are more important than horizontal ones because of temperature and pressure changes in this direction are much larger than similar distances in the horizontal direction.

17. Pressure and Altitude Station pressure: Local Reading Local Values must be corrected for elevation and reduced to sea level for comparison.

18. Pressure Lapse Rate In an average standard atmosphere with a lapse rate of ~ 6.5 o C/1000 m, atmospheric pressure decreases are approximated by ~ 100 mb/1000 m. Station pressure at its elevation is then corrected to sea level with this standard gradient. Example:Station Pressure 894 mb, Elevation +1100 m Equivalent Sea Level Pressure = 894 + 1100*(100/1000) = 1004 mb

19. Pressure Gradients  P/  d = Pressure Gradient (Change in Pressure/Distance) Example:  P/  z = -  g  P = change in pressure  z = change in altitude  = density (air) g = gravity (9.8 m/s 2 ) Pressure Gradient Force (PGF): Force due to pressure differences, and the cause of air movement (winds).

20. Pressure Gradient Force F = m a, acceleration a = F/m =  v/  t P = F/Apressure  = m/Vdensity acceleration a = F/m = -(1/  )*  P/  d Wind is accelerated perpendicular to isobars from High to Low pressure. Wind velocity ~ (1/  *  P/  d

21. Isobars Contours of Constant Pressure Topographical Analogy: Just as close topographic lines indicate steep terrain, close isobars mean steep pressure gradient, large pressure force and therefore strong winds. SURFACE MAP

22. Pressure Gradient Force (PGF) Air moves from High pressure to Low pressure. The force provided by the pressure difference is the pressure gradient force. Force is directed perpendicular to the isobars from Highs to Lows.

23. Isobar Levels Standard isobaric levels Each isobaric level picks out a general altitude to study. These are constant altitude charts, similar to the surface map.

24. Fig. 8-12, p. 199

25. Isobar Levels

26. Bernoulli’s Equation Recall that WORK = change in both Potential Energy and Kinetic Energy. W = Force * Distance = P A x = P V Initially, P 1, v 1, x 1, h 1, A 1 Finally, P 2, v 2, x 2, h 2, A 2 h2h2

27. Bernoulli’s Equation  KE = 1/2 m (v 2 -v 1 ) 2  PE = mg (h 2 -h 1 ) Initially, P 1, v 1, x 1, h 1, A 1 Finally, P 2, v 2, x 2, h 2, A 2 h2h2

28. Bernoulli’s Equation  KE = 1/2 m (v 2 -v 1 ) 2  PE = mg (h 2 -h 1 )  W =  KE +  PE Leads to the equation of continuity... P + 1/2  v 2 +  g h = constant The sum of pressure, kinetic energy per volume and potential energy per volume have the same value along all points in a streamline.

29. Bernoulli’s Principle Pressure is lower for faster steady flow than for slower Airplane Wings Higher v, Lower P Lower v, Higher P

30. Wind Speed Vectors Vectors are arrows representing wind direction and speed.

31. Bernoulli Winds

32. Ridge and Trough Wavelike patterns of constant pressure surfaces reflect pressure changes due to air temperature differences.

33. Constant Pressure Surfaces At what height do you experience a Pressure of 500 mb?

34. Density and Pressure The warmer air column is less dense. At the same height, the pressure is higher in the warmer air column.

35. Latitude Pressure Gradient At the same altitude, the pressure is higher on average in the warmer (i.e. lower) latitudes.

36. Temperature PGF versus Altitude

37. Air Forces Force = Pressure Gradient Force + Coriolis + Friction For any fixed height, the pressure decreases toward cooler latitudes. Isobars for a fixed altitude show this decrease to the North (South) in the Northern (Southern) Hemisphere. In the absence of rotation, air (in the upper level) would tend to flow from the equator toward the poles.

38. Global Circulation In the absence of rotation, air would tend to flow from the equator toward the poles. Hot, less dense air rising at the equator, becomes denser as it cools and descends at the poles, traveling back to tropical areas to heat up again.

39. Coriolis

40. Outsider’s view Insider’s view From http://www.windpower.dk/tour/wres/coriolis.htm Outsider’s view Insider’s view

41. A. Force: due to a rotating frame of reference. B. Objects moving in a straight line with respect to the stars, will experience an apparent deflection to the RIGHT in the N.Hemisphere and an apparent deflection to the LEFT in the S. Hemisphere. C. The Coriolis force is strongest at the poles and zero at the equator. D. The Coriolis force is proportional to the speed. E. The "force" affects the direction NOT the speed. But since velocity is a vector, with both direction and speed, the velocity change is the same as acceleration and the Coriolis force is the culprit. Coriolis FORCE

42. Relative to a carousel, someone walking on a carousel moves in a straight line with respect to the fixtures. Relative to others and equipment on the ground, the person moves in an arc as if affected by a force. The (horizontal) Coriolis Force F c = 2  v sin  v = wind speed  = angular velocity (earth rotation, 360 degrees/24 hours)  = latitude (sin 0 = 0.0 equator, sin 90 = 1.0 poles) Coriolis Force Equation

43. Coriolis Deflection F c = 2  v sin  Increase of the Coriolis Force with wind speed.

44. Pressure Gradient Force Air flows from high pressure to low pressure, so on average, from the equators to the poles.

45. Geostrophic flow Remember that the coriolis force depends upon velocity. As air is accelerated by the PGF its speed increases and the coriolis deflection grows. Equilibrium is reached when the PGF and coriolis effect are equal.

46. PGF = Coriolis 1/  *  P/  d = 2  v sin  The wind velocityv =  P/  d * (1/  2  sin  ) Utilizing the coriolis parameter f c = 2  sin  The wind velocity v = (1/  f c )*  P/  d = v g This is known as the geostrophic wind equation.

48. Geostrophic flow With the inclusion of the Coriolis Force, air flows parallel to isobars of constant pressure.

49. Westerlies At mid latitudes, air moving from S. to N. in the northern hemisphere flows from west to east.

51. Friction Effect Friction retards wind speed near the surface due to topography, lowering the coriolis force. Therefore, wind direction is altered from parallel to isobars.

52. Cyclonic Flow

53. Low Pressure Cyclonic Winds

54. High Pressure Cyclonic Winds

55. Isobar Surface Map

56. Cyclone Anticyclone Circulations

57. Summary Pressure = Force / Area Forces on Air –Pressure Gradient Force –Coriolis Force –Friction Isobar Charts Global Circulation Pattern (3 cell model) Cyclone and Anticyclone Circulations