Lectures_7 - Winds and Atmospheric Forces Fig. 6-CO, p.140...

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Fig. 6-CO, p.140 Winds and Atmospheric Forces
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Atmospheric Forces Gravity (already encountered) Pressure Gradient (PG) force Centripetal/centrifugal force Coriolis Force/acceleration Apply to diagnosis of wind systems
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Fig. 6-1, p.142 From the text: A model of the atmosphere where air density remains constant with height. The air pressure at the surface is related to the number of molecules above. When air of the same temperature is stuffed into the column, the surface air pressure rises. When air is removed from the column, the surface pressure falls.
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Fig. 6-2a, p.143 No pressure difference – no horizontal flow
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Fig. 6-2b, p.143 Same pressure at surface but pressure difference aloft
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Fig. 6-2c, p.143 Pressure difference aloft cause horizontal flow Which in turn causes pressure difference at surface And return surface flow
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Fig. 6-9, p.150 Excess pressure or pressure gradient causes flow from high pressure to low pressure
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Fig. 6-3, p.145 Standard MSLP Pressure Extremes
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Fig. 6-5, p.146
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Fig. 6-7, p.147 Extrapolating pressure to constant surface
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Y X P(x) P(x+ x) Consider a cube of air with sides X and Y, area A, mass m Force on face at X in the X-direction P(X).A Newtons to RHS Force of face at X+ X P(X+ X).A = (P + P).A to LHS Net force to RHS = P(X).A – (P + P).A = - P.A Pressure Gradient Force N.B. Each Face is a constant Pressure surface and the force is normal to it.
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Pressure Gradient Force Force acts on mass m = A. x. ρ (air) kg • Force/mass = F x = - ( P.A)/(A. x. ρ ) N.B. this is not a volume force like gravity Also for the force on the cube in the y- direction Force is normal to constant pressure surface ρ = - N/kg x P F X = - N/kg Y P F Y
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Pressure gradient Force N.B. Both Fx and Fy act from Hi to Lo pressure In general the force is normal to the isobars Vector ρ = - = - 1 1 ˆ F P P F n n n 101 kPa 100 kPa 99 kPa n
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Pressure Gradient Force (PGF) Isobars Map Scale Map: needs scale/direction Magnitude of PGF/unit mass = 1/ ρ∆ P/ n 2200 P = pressure between isobars (Pascals) 2200 n = shortest distance between isobars (m) 2200 ρ = air density (kg/m 3 ) N kPa 99 100 98 101 P n 1° lat =110 km P = 1kPa = 1000 Pa n = 0.8° 88 km = 88x1000 m
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Atmospheric Forces: Centrifugal/petal Two ways of viewing this situation Fixed inertial frame - acceleration Frame moving with the object – forces in equilibrium Forces “become” acceleration and vice versa An object, mass m, rotates at the end of a string in tension. Tension causes acceleration m a = T which is perpendicular to the velocity viewed from the fixed frame the object Viewed from frame moving with object the centrifugal force T V V a T Fcent
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R V= xR F cent = x xR Atmospheric Forces: Centrifugal/petal Rotation with radius R Angular speed, Centrifugal force is – F cent = R 2 = V 2 /R magnitude V = xR F cent = x xR vector
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This note was uploaded on 06/05/2010 for the course EATS 1011 taught by Professor Johnm during the Winter '10 term at York University.

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Lectures_7 - Winds and Atmospheric Forces Fig. 6-CO, p.140...

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