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NATS 101 Lecture 14 Air Pressure

NATS 101 Lecture 14 Air Pressure. What is Air Pressure?. Pressure = Force/Area What is a Force? It’s like a push/shove In an air filled container, pressure is due to molecules pushing the sides outward by recoiling off them. Recoil Force. Recoil Force. Air Pressure.

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NATS 101 Lecture 14 Air Pressure

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  1. NATS 101Lecture 14Air Pressure

  2. What is Air Pressure? Pressure = Force/Area What is a Force? It’s like a push/shove In an air filled container, pressure is due to molecules pushing the sides outward by recoiling off them Recoil Force

  3. Recoil Force Air Pressure Concept applies to an “air parcel” surrounded by more air parcels, but molecules create pressure through rebounding off air molecules in other neighboring parcels

  4. Recoil Force Air Pressure At any point, pressure is the same in all directions But pressure can vary from one point to another point

  5. Higher density at the same temperature creates higher pressure by more collisions among molecules of average same speed Higher temperatures at the same density creates higher pressure by collisions amongst faster moving molecules

  6. Ideal Gas Law • Relation between pressure, temperature and density is quantified by the Ideal Gas Law P(mb) = constant  (kg/m3)  T(K) • Where P is pressure in millibars • Where  is density in kilograms/(meter)3 • Where T is temperature in Kelvin

  7. Ideal Gas Law • Ideal Gas Law is complex P(mb) = constant  (kg/m3)  T(K) P(mb) = 2.87  (kg/m3)  T(K) • If you change one variable, the other two will change. It is easiest to understand the concept if one variable is held constant while varying the other two

  8. Ideal Gas Law P = constant    T (constant) With T constant, Ideal Gas Law reduces to  P varies with  Denser air has a higher pressure than less dense air at the same temperature Why? You give the physical reason!

  9. Ideal Gas Law P = constant   (constant)  T With  constant, Ideal Gas Law reduces to  P varies with T Warmer air has a higher pressure than colder air at the same density Why? You answer the underlying physics!

  10. Ideal Gas Law P (constant) = constant    T With P constant, Ideal Gas Law reduces to T varies with 1/ Colder air is more dense ( big, 1/ small) than warmer air at the same pressure Why? Again, you reason the mechanism!

  11. Summary • Ideal Gas Law Relates Temperature-Density-Pressure

  12. Pressure-Temperature-Density Pressure Decreases with height at same rate in air of same temperature Isobaric Surfaces Slopes are horizontal 300 mb 400 mb 500 mb 9.0 km 9.0 km 600 mb 700 mb 800 mb 900 mb 1000 mb Minneapolis Houston

  13. Pressure-Temperature-Density WARM Pressure (vertical scale highly distorted) Decreases more rapidly with height in cold air than in warm air Isobaric surfaces will slope downward toward cold air Slope increases with height to tropopause, near 300 mb in winter 300 mb COLD 400 mb 500 mb 9.5 km 600 mb 700 mb 8.5 km 800 mb 900 mb 1000 mb Minneapolis Houston

  14. Pressure-Temperature-Density WARM Pressure Higher along horizontal red line in warm air than in cold air Pressure difference is a non-zero force Pressure Gradient Force or PGF (red arrow) Air will accelerate from column 2 towards 1 Pressure falls at bottom of column 2, rises at 1 Animation 300 mb COLD 400 mb 500 mb H L PGF 9.5 km 600 mb 700 mb 8.5 km 800 mb 900 mb 1000 mb H L PGF Minneapolis Houston SFC pressure rises SFC pressure falls

  15. Summary • Ideal Gas Law Implies Pressure decreases more rapidly with height in cold air than in warm air. • Consequently….. Horizontal temperature differences lead to horizontal pressure differences! And horizontal pressure differences lead to air motion…or the wind!

  16. Review: Pressure-Height Remember • Pressure falls very rapidly with height near sea-level 3,000 m 701 mb 2,500 m 747 mb 2,000 m 795 mb 1,500 m 846 mb 1,000 m 899 mb 500 m 955 mb 0 m 1013 mb 1 mb per 10 m height Consequently……….Vertical pressure changes from differences in station elevation dominate horizontal changes

  17. Station Pressure Ahrens, Fig. 6.7 Pressure is recorded at stations with different altitudes Station pressure differences reflect altitude differences Wind is forced by horizontal pressure differences Horizontal pressure variations are 1 mb per 100 kmAdjust station pressures to one standard level: Mean Sea Level

  18. Reduction to Sea-Level-Pressure Ahrens, Fig. 6.7 Station pressures are adjusted toSea Level PressureMake altitude correction of 1 mb per 10 m elevation

  19. Correction for Tucson Elevation of Tucson AZ is ~800 m Station pressure at Tucson runs ~930 mb So SLP for Tucson would be SLP = 930 mb + (1 mb / 10 m)800 m SLP = 930 mb + 80 mb = 1010 mb

  20. Correction for Denver Elevation of Denver CO is ~1600 m Station pressure at Denver runs ~850 mb So SLP for Denver would be SLP = 850 mb + (1 mb / 10 m)1600 m SLP = 850 mb + 160 mb = 1010 mb Actual pressure corrections take into account temperature and pressure-height variations, but 1 mb / 10 m is a good approximation

  21. You Try at Home for Phoenix Elevation of Phoenix AZ is ~340 m Assume the station pressure at Phoenix was ~977 mb at 3pm yesterday So SLP for Phoenix would be?

  22. Sea Level Pressure Values Ahrens, Fig. 6.3

  23. Summary • Because horizontal pressure differences are the force that drives the wind Station pressures are adjusted to one standard level…Mean Sea Level…to remove the dominating impact of different elevations on pressure change

  24. PGF Ahrens, Fig. 6.7

  25. Key Points for Today • Air Pressure Force / Area (Recorded with Barometer) • Ideal Gas Law Relates Temperature, Density and Pressure • Pressure Changes with Height Decreases more rapidly in cold air than warm • Station Pressure Reduced to Sea Level Pressure

  26. Assignment • Reading - Ahrens pg 148-149 include Focus on Special Topic: Isobaric Maps • Problems - 6.9, 6.10

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