Meteorology for Air Pollution Control Engineers7

Meteorology for Air Pollution Control Engineers7 - of 1...

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1 Example 3 n Estimate the Coriolis acceleration for a body moving 10 ft/s at  40o North latitude. n Solution: Using the earth’s anaular velocity from Example 1, we find  Coriolis acceleration= 2 × 10ft/s × 7.27 × 10-5/s × sin 40o = 9.35 × 10-4 ft/s2 = 2.85 × ×10-4 m/s2    # n It’s 2.9 × 10-5 as large as the acceleration of gravity.

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2 n The principal accelerating forces causing or retarding  horizontal flow in the atmosphere are the Coriolis force,  pressure gradient forces, and frictional resistance at the  surface of the earth.
3 Example 4 n Estimate the acceleration of the air caused by a pressure gradient

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Unformatted text preview: of 1 mb/100 km. (1 bar = 105 Pa = 0.9872 atm) 1 mb/100 km = 1 Pa/km n Solution: Apply Newton’s law to 1 cubic km of air (a cube with edge length = x = 1 km) and use the standard sea-level air density, finding 4 n Pressure acceleration n This is a typical atmospheric pressure gradient. It is about three times the Coriolis acceleration, but only about 8 × 10-5 times the acceleration of gravity. ( 29 ( 29 ( 29 3 2 2 3 1 1000 1 1.21 / 4 2 3 2 / / / / 8.3 10 / 2.7 10 / # kg Pa km m km kg m S m Pa F m A p V x p x p x m s ft s ρ × × ×--= = ∆ = ∆ = ∆ = = × = ×...
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Meteorology for Air Pollution Control Engineers7 - of 1...

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