# air pressure nats 101 lecture 14 air pressure. recoil force what is air pressure? pressure =...

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• Slide 1
• Air Pressure NATS 101 Lecture 14 Air Pressure
• Slide 2
• Recoil Force What is Air Pressure? Pressure = Force/Area What is a Force? Its like a push/shove In an air filled container, pressure is due to molecules pushing the sides outward by recoiling off them
• Slide 3
• 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 Recoil Force
• Slide 4
• Air Pressure At any point, pressure is the same in all directions But pressure can vary from one point to another point Recoil Force
• Slide 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
• Slide 6
• Ideal Gas Law Relation between pressure, temperature and density is quantified by the Ideal Gas Law P(mb) = constant x d(kg/m 3 ) x T(K) Where P is pressure in millibars Where d is density in kilograms/(meter) 3 Where T is temperature in Kelvin
• Slide 7
• Ideal Gas Law Ideal Gas Law is complex P(mb) = constant x d(kg/m 3 ) x T(K) P(mb) = 2.87 x d(kg/m 3 ) x 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
• Slide 8
• Ideal Gas Law P = constant x d x T (constant) With T constant, Ideal Gas Law reduces to P varies with d Boyle's Law Denser air has a higher pressure than less dense air at the same temperature Why? You give the physical reason!
• Slide 9
• Ideal Gas Law P = constant x d (constant) x T With d constant, Ideal Gas Law reduces to P varies with T Charles's Law Warmer air has a higher pressure than colder air at the same density Why? You answer the underlying physics!
• Slide 10
• Ideal Gas Law P (constant) = constant x d x T With P constant, Ideal Gas Law reduces to T varies with 1/d Colder air is more dense (d big, 1/d small) than warmer air at the same pressure Why? Again, you reason the mechanism!
• Slide 11
• Summary Ideal Gas Law Relates Temperature-Density-Pressure
• Slide 12
• Pressure-Temperature-Density 9.0 km 300 mb 1000 mb 400 mb 500 mb 600 mb 700 mb 800 mb 900 mb MinneapolisHouston 9.0 km Pressure Decreases with height at same rate in air of same temperature Constant Pressure (Isobaric) Surfaces Slopes are horizontal
• Slide 13
• Pressure-Temperature-Density 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 8.5 km 9.5 km 300 mb 1000 mb 400 mb 500 mb 600 mb 700 mb 800 mb 900 mb MinneapolisHouston COLD WARM
• Slide 14
• Pressure-Temperature-Density 8.5 km 9.5 km 300 mb 1000 mb 400 mb 500 mb 600 mb 700 mb 800 mb 900 mb MinneapolisHouston H L LH 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 SFC pressure risesSFC pressure falls PGF COLD WARM
• Slide 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 motionor the wind!
• Slide 16
• N. Pacific Pressure Analysis (isobars every 4 mb) Pressure varies by 1 mb per 100 km horizontally or 0.0001 mb per 10 m 2000 km
• Slide 17
• Review: Pressure-Height Remember Pressure falls very rapidly with height near sea-level 3,000 m 701 mb 2,500 m747 mb 2,000 m 795 mb 1,500 m846 mb 1,000 m899 mb 500 m955 mb 0 m1013 mb 1 mb per 10 m height Consequently. Vertical pressure changes from differences in station elevation dominate horizontal changes
• Slide 18
• Station Pressure Pressure is recorded at stations with different altitudes Station pressure differences reflect altitude differences Wind is forced by horizontal pressure differences Since horizontal pressure variations are 1 mb per 100 km We must adjust station pressures to one standard level: Mean Sea Level Ahrens, Fig. 6.7
• Slide 19
• Reduction to Sea-Level-Pressure Sea Level Pressure Station pressures are adjusted to Sea Level Pressure Make altitude correction of 1 mb per 10 m elevation Ahrens, Fig. 6.7
• Slide 20
• 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) x 800 m SLP = 930 mb + 80 mb = 1010 mb
• Slide 21
• 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) x 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
• Slide 22
• 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?
• Slide 23
• Sea Level Pressure Values Ahrens, Fig. 6.3 (October, 2005) Wilma 882 mb (26.04 in.)
• Slide 24
• Summary Because horizontal pressure differences are the force that drives the wind Station pressures are adjusted to one standard levelMean Sea Levelto remove the dominating impact of different elevations on pressure change
• Slide 25
• Ahrens, Fig. 6.7 PGF
• Slide 26
• Key Points 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 Mean-Sea-Level to Mitigate the Dominate Impact of Altitude on Pressure Change
• Slide 27
• NATS 101 Lecture 15 Surface and Upper-Air Maps
• Slide 28
• Supplemental References for Todays Lecture Gedzelman, S. D., 1980: The Science and Wonders of the Atmosphere. 535 pp. John-Wiley & Sons. (ISBN 0-471-02972-6)
• Slide 29
• Summary Because horizontal pressure differences are the force that drives the wind Station pressures are adjusted to one standard levelMean Sea Levelto mitigate the impact of different elevations on pressure
• Slide 30
• Ahrens, Fig. 6.7 PGF
• Slide 31
• Local Example The station pressure at PHX is ~977 mb. The station pressure at TUS is ~932 mb. Which station has that higher SLP?
• Slide 32
• Correction for Phoenix Elevation of PHX Airport is ~340 m Station pressure at PHX was ~977 mb So, SLP for PHX would be SLP = 977 mb + (1 mb / 10 m) x 340 m SLP = 977 mb + 34 mb = 1011 mb
• Slide 33
• Correction for Tucson Elevation of TUS Airport is ~800 m Station pressure at TUS was ~932 mb So, SLP for TUS would be SLP = 932 mb + (1 mb / 10 m) x 800 m SLP = 932 mb + 80 mb = 1012 mb The SLP was higher in TUS than PHX
• Slide 34
• Surface Maps Pressure reduced to Mean Sea Level is plotted and analyzed for surface maps. Estimated from station pressures Actual surface observations for other weather elements (e.g. temperatures, dew points, winds, etc.) are plotted on surface maps. NCEP/HPC Daily Weather Map
• Slide 35
• Assignment Reading - Ahrens pg 148-157 include Focus on Special Topic: Isobaric Maps Problems - 6.9, 6.10, 6.12, 6.13, 6.17, 6.19, 6.22

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