Hyeon-Jun Kim Sang-Yong Park Hwa-Jeong Shin Eun-Young Kim

The hydrostatic Equation Geopotential Scale Height and the Hypsometric Equation Thickness and Heights of Constant Pressure Surfaces Reduction of Pressure to Sea Level

3.2.1 The Hydrostatic Equation Hydrostatic balance The net upward on the slab is equal to the downward force on the slab. Hydrostatic equation ∂p/ ∂z = -gρ

Ex) Vertical column of air with unit hori-zonal cross-sectional area (hydrostatic balance) The net upward force on slab The downward force slab -δp = gρδz in the limit as δz → 0

∂p/ ∂z = -gρ change ρ = 1/a gdz = -αdp ex) If the pressure at height z is p(z)

p(∞)=0, - pressure become small in higher and higher. so - The pressure at height z is equal to the vertical column of unit cross-sectional area lying above that level.

the gravitational potential per unit mass J kg -1 or m 2 s -2 dФ ≡ gdz=-αdp The geopotential Ф(z) at height z is Geopotential

the geopotential height Z is (g 0 =9.81ms -2 ) Used as the vertical coordinate in most atmospheric applications in which energy plays an important role

Eliminating ρ, The (geopotential) thickness of the layer between pressure levels p 1 and p 2

3.2.2 Scale Height For an isothermal atmosphere, if the virtual temperature correction is neglected

A scale height is a term often used in scientific contexts for a distance over which a quantity decreases by a factor of e. It is usually denoted by the capital letter H.

Ex 3.2 the number density of oxygen atoms the number density of hydrogen atoms = 10 5 (200km) 2000 K

1 2

Hypsometric Equation if we define with respect to p as shown in this fig. Hypsometric equation

Ex 3.3 Calculate the geopotential height of the 1000 hPa pressure surface when the pressure at sea level is 1014 hPa, The scale height of the atmosphere may be taken as 8 km. ln(l + x) ~ x for x <<1 = 112 m

)ln() p p g TR p p HZZ vd  Thickness Thickness : The difference in height between two atmospheric pressure levels Hypsometric equation Thickness Mean virtual temperature  Geopotential height Earth 2 zz 1

Ex 3.4 Calculate the thickness of the layer between the 1000 and 500hPa pressure surfaces Surfaces (a) tropics = 15 ℃ =288K(tropics) (b)polar regions =-40 ℃ =233K(polar regions) TvTv TvTv Solution from hypsometric eq. Therefore, (a) ΔZ=5846m (b) ΔZ=4730m ) 2 1 p p ln() g TR p p HZZ vd 

a.Hurricane –Center of a storm is warmer than its surroundings. –Intensity of the storm must decrease with height Heights of Constant Pressure Surfaces b.Upper level lows –Below that level must be cold core –Above that level must be warm core Greatest intensity

3.2.4 Reduction of pressure to sea level Sea level - Point of standard ground height and the depth of water the seafloor. - To isolate that part of the pressure field use common reference level. - Z 0 = 0 so

− p 0 = pg exp(Zg/H) = pg exp(Zgg 0 /RdTv) * if Zg/H <<1, approximated 1+ Zg/H (because e to 0 is 1, e to 1 is 2.7xx) − p 0 - pg = pg(Zg/H) = pg(Zgg 0 /RdTv) − pg near 1000 and H near 8000m − p 0 - pg = 1000(Zg/8000m) The pressure decreases by about 1hPa for every 8m of vertical ascent.

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