Scale analysis of the vertical equation of motion:

Transcription

1 Scale analysis of the vertical equation of motion: As we did with the hz eqns, we do for the vertical to estimate the order of magnitude of Dw/ we take the largest of the terms, Dw -- W/T h UW/L W 2 /H w - + u w - + v- w + w w t x y z z-equation Dw/ -2Ωucosφ -(u 2 +v 2 )/a ( ρ) p z -g scaling WU/L f 0 U U 2 /a δp v /(ρh) g value Thus we see that the atmosphere, to the first order, is hydrostatic, p ρg z Even though the atmosphere technically is not at rest, the hydrotatic apx is not a bad estimate. However, the story does not end here, there is more to this than meets the eye in that g and ( ρ) p z nearly cancel each other out (i.e. the -0 scenario). Thus, slight differences between two relatively large terms are responsible for the vertical acceleration. In order to examine these terms in more detail, we introduce the concept of mean and perturbation quantities. Mass Conservation The mass of an air parcel is conserved [no matter how it moves around, deforms, stretches, etc.] Let V be a parcel's volume and ρ it's density. m ρv is the parcel's mass. Mass conservation can be written in many equivalent Lagrangian forms: m const [but dif parcels have dif masses, i.e. dif consts] ρv const ρ(t)v(t) const [ If V increases, ρ must decrease] Dm/ 0 D(ρV)/ 0

2 Use product rule: ρ DV -- V-- Dρ ρ DV -- V Dρ + -- ρv 0 -- DV Dρ -- 0 V ρ These are all equivalent Lagrangian forms. Eulerian forms of mass conservation First lets focus on /V DV/ -- we'll be able to relate it to spatial changes in the wind field. Consider an infinitesimal box-shaped air parcel: z x y Vδxδyδz Thus, from our box we have: air δx δy δz -- DV -- V -- D ( δxδyδz) δxδyδz -- δyδzδx D δxδyδz + δxδzδy D + δxδyδz D δx D D + δy δz D δx δy δz Let s take a look at the time rate of change of the line segment δx: x A δx(t0) x B x A x B δx(δt) Note the following: Translation or rotation of line segment does NOT change the length of the segment! In an infinitesimal time (δt), the v and w components can rotate and translate the line segment but cannot elongate (stretch) it.

3 y x translate tδt t0 rotate elongate u can translate and elongate the segment δx Therefore we have: δx(0) x B x A [initial length] δx(δt) x B x A [length a short time later] Taylor series approximation of x A for small t: dx 0 A x A x A + ---δt + HOT x dt A + u A δt Similarly, dx B 0 x B x B + ---δt + HOT x dt B + u B δt Subtracting, we get x B x A x B x A + ( u B u A )δt δx( δt) δxt ( 0) + ( u B u A )δt Also apply Taylor series for u B, u ~0 u B u A u + ( x x B x A ) + HOT u A + δx ( 0) x Plugging in the expansion about u B into δx(δt) we have u B -u A u δx( δt) δx( 0) + δx ( 0)δt x rearrange (solve for u x )

4 u x δx( δt) δx( 0) δx( 0)δt lim δx( δt) δx( 0) δt 0 δx( 0)δt δx D ---- ( δx) we can show similarly v y ---- Dδy w, and - δy z ---- Dδz ---- δz Hence, we can now relate the three-dimensional wind to the change in volume -- DV -- V u v w u x y z This is 3-D divergence! If u >0 (divergence) air parcel expands (i.e. volume V increases) Now that we have related the change in volume to the velocity, we can rewrite the Lagrangian form of mass conservation as: -- DV Dρ + u Dρ 0 Mass Conservation V ρ ρ If u >0 then Dρ/<0 (which is consistent with above, i.e. divergence and increasing volume and therefore decreasing density). The above expression is a hybrid - part Eularian and part Lagrangian. We can get rid of the mix by multiplying by ρ and expanding the total derivative, ρ ρ + u ρ + ρ u 0 + ρu t t 0 (flux form) where I have used the product rule to condense the last two terms of the first equation above. These two expressions have no total derivatives in them - they are purely Eularian frame. A couple of points... Note that the horizontal divergence provides information on how the area changes but not how the volume changes. For example

5 H u u v + δx D D + δy - D ( δxδy) x y δx δy δxδy - DA H - A H where A H is the horizontal area. CASE I: If u x, v y > 0, we have DA H / > 0 and the area grows δy(0) δy(δt) δx(0) δx(δt) CASE II: If u x 0, and v y < 0, we have compression along the y axis and the area shrinks δy(0) δx(0) δy(δt) δx(δt) CASE III: If u x > 0, v y > 0, and w z > 0, the parcel expands in x, y, and z directions (i.e. the volume increases) and the horizontal area increases. CASE IV: If u x s, v y 2s, and w z 3s, then we have DA H / < 0 (shrinking horizontal area) but growing volume! Thus parcel is stretching in the x and z-directions, and compressing in the y direction. STUDENTS: Show (in 3-D) that ρu ρ u + u ρ

6 Scaling the Mass Conservation Equation As with the vertical equation of motion, we decompose the density into base state (mean) and perturbation quantities, ---- ( ρ + ρ t 0 ) + u ρ ( + ρ 0 ) + ( ρ + ρ 0 ) u 0 where ρ( x, y, z, t) ρ 0 ( z) + ρ ( xyzt,,, ). We know that ρ 0 is a f(z) only rewritten ( ρ 0 t, ρ 0 x, ρ 0 y 0), thus the above equation can be ---- ρ u ρ w dρ ( ρ + ρ t dz 0 ) u 0 multiplying by /ρ 0 we have ---- ρ -- + u ρ ρ 0 t ---- w dρ small ρ u + u 0 ρ 0 dz ρ 0 when comparing the last two terms we know that ρ ρ 0 «0, thus the 4th term is much smaller than the 5th term above. We now have what Holton has, namely ---- ρ -- + u ρ ρ 0 t ---- w dρ u 0 ρ 0 dz