3. Rotational Dynamics Flashcards

Question

Coriolis Dominated Drift

Answer 1

-drift satisfies the pure coriolis equations with zero acceleration (zero inertia) =\> y = constant x = at/2Ω = drift

Answer 2

- for t \>\> 1/Ω, drift is dominant - cf geostrophic balance in fluid dynamics

Answer 3

ρDu\_/Dt = -∇p + ∇(1/2 ρΩ²r²) - 2ρΩ\_xu\_ + F\_ -taking F\_=0 =\> ρDu\_/Dt = -∇(p - 1/2 ρΩ²r²) - 2ρΩ\_xu\_ ∇.u\_ = 0

Answer 4

- for certain situations, the predictions of the governing equations are the same with or without Fu\_ - this is the case for closed boundries where the reaction force from the solid boundary balances the centrifugal force

Answer 5

-for rotating fulids, p is replaced in the governing equation by P, where: P = p - 1/2 ρΩ²r²

Answer 6

- imagine a circular room rotating with angular velocity Ω - inside the room is an object rotating with angular velocity ζ/2 - an observer in the rotating reference frame observes the rotation vector of the object as ζ/2 - an inertial observer outside of the rotating room observers the rotation vector of the object as ζ/2 + Ω, which is the actual rate of rotation - the angular momentum of the object as observed by the inertial observer needs to be conserved whereas the angular momentum as observed by the rotational observer is not conserved

Answer 7

ζ\_ = ∇\_ x u\_

Answer 8

q\_ = ζ\_ + 2Ω\_

Answer 9

Dq\_/Dt = q\_ . ∇\_u\_ - where Dq\_/Dt is the rate of change of q\_ for a fluid element - and q\_ . ∇\_u\_ represents vortex stretching

Answer 10

- say, q\_=q ez^ - then the vorticity equation becomes: q\_ . ∇\_u\_ = q ∂w/∂z = Dq/Dt -rate of change of absolute vorticity = rate of extension

Answer 11

-fluid mechanical analogue of inertial oscillations for particles (also called inertial oscsillations) ∂u/∂t = - 1/ρ ∂p/∂x - 2Ωv ∂v/∂t = - 1/ρ ∂p/∂y + 2Ωu ∂w/∂t = - 1/ρ ∂p/∂z -and incompressibility: ∂u/∂x + ∂v/∂y + ∂w/∂z = 0

Answer 12

inertia ~ ρU²/L Coriolis ~ ρΩU -take the ratio: inertia / Coriolis = U/ΩL = Ro -where Ro is the Rossby Number

Answer 13

Ro \<\< 1 =\> Coriolis dominant, geostrophic flow, inertial oscillations negligible, planetary scales Ro \>\> 1 =\> Inertia dominant, cyclostrophic flow, everyday sclaes -e.g. for the ocean Coriolis is dominant over inertia, for the atmosphere Coriolis is also dominant but less so and for Jupiter the result is similar

Answer 14

-Coriolis dominated fluid dynamics so inertial term, ρDu\_/Dt, can be set to 0 =\> 2ρ Ω\_ x u\_ = -∇\_P + F\_ -this gives three equations, x, y, z, as well as the incompressibility condition

Answer 15

- if the pressure field is known, then u and v can be calculated directly - the horizontal divergence (∂u/∂x + ∂v/∂y) = 0, so using incompressibility: ∂w/∂z = -(∂u/∂x + ∂v/∂y) = 0 - w is independent of z - also horizontal divergence = 0 =\> existence of streamfunction Ψ such that: u = -∂Ψ/∂y, v = ∂Ψ/∂x -where Ψ=Ψ(x,y)

Answer 16

-using the governing equations, Ψ = P / 2Ωρ =\> P is a streamfunction with constant -isobars are streamlilnes since Ψ=const. \<=\> P=const.

Answer 17

- when pressure is low, you get anticlockwise rotation, a cyclonic weather system - when pressure is high, you get clockwise rotation, an anticyclonic weather system

Answer 18

- would expect greatest Coriolils effect at poles and zero at equator - this is generally true but dynamics at the pole are dominated by other mechanisms

Answer 19

- for a geostrophic system, ∂w/∂z=0 - calculating ∂u/∂z from the governing equations =\> ∂u/∂z = 0 - and similarly, ∂v/∂y=0 - a geostrophich flow does not depend on z

Answer 20

- by the Taylor-Proudman theorem, a geostrophic flow does not depende on z - so if there is an obstacle at ground level z=0, for example an island then the motion of the flow around this obstacle will be replicated at every value of z as if the flow is movin around a virtual obstacle - the resulting structures are called Taylor columns

Answer 21

- even if flow is geostrophic in the upper atmosphere, further down the ground exerts friction - model this by incorporating a viscous stress due to vertical shear (cf. inertial boundary layer, Blassius)

Answer 22

0 = -∇\_P - 2ρΩ\_ x u\_ + ν∂²u\_/∂z² - gives three equations: - 2Ωv = -1/ρ ∂p/∂x + ν∂²u\_/∂z² 2Ωu = -1/ρ ∂p/∂y + ν∂²v\_/∂z² ∂p/∂z = -ρg -boundary conditions: u ~ uo(x,y) and v~vo(x,y) as z-\>∞, i.e. known geostrophic flow in the upper atmosphere u=v=0 at z=0, no-slipi boundary condition

Answer 23

-integrate the z equation for: P = -ρgz + po(x,y) - substitute into each of x and y and apply the geostrophic limit boundary condition - let U=u+iv, this gives a second order equation for U - solve with particular integral U=Uo (Uo=uo+ivo) and complimentary function (U=Be^[-z√[2Ωi/ν]] - plotting v against u gives an Ekman spiral

Answer 24

-using incompressibility: ∂w/∂z = - (∂u/∂x + ∂v/∂y) -integrate with respect to z =\> w = ζo(x,y) [1 - e^(z/L) (sin(z/L)+cos(z/L))] - where ζo(x,y) is the relative vorticity of the geostrophic flow at z-\>∞ - in the limit z-\>∞, w ~ Lζo/2 ≠ 0 - so the upper atmosphere is moving up and down w\>0 if ζo\>0 =\> low pressure weather system w\<0 if ζo\<0 =\> high pressure weather system

Answer 25

- when there is low pressure, moisture is driven upwards to colder air where is condenses forming clouds and possibly precipitation - when there is high pressure, moisture staus at ground level, clear skies

Answer 26

2Ω\_ = 2Ωcosθ ey^ + 2Ωsinθ ez^ where 2Ωcosθ = γ and 2Ωsinθ = f

Answer 27

Du/Dt - fv = -1/ρ ∂p/∂x Dv/Dt + fu = -1/ρ ∂p/∂y Dw/Dt = -1/ρ ∂p/∂z - g

Answer 28

- often interested in a single zone, θ=θo (i.e. a latitude) so let θ^\<\<θo and θ = θo + θ^ - then f = 2Ωsinθ = 2Ωsin(θo + θ^) - use double angle formula and linearise =\> f = 2Ωsinθo + 2Ωθ^cosθo f = f + βy -where θ^ = y/R

Answer 29

fo\>0 in the Northern hemisphere fo\<0 in the Southern hemisphere

Answer 30

- if f≈fo , f-plane approx. - if f≈fo+βy , β-plane approx - important for Rossby waves

Answer 31

q\_ = f\_ + ζ\_ - where q\_ is the absolute vorticity - can show: Dq\_/Dt = q\_ . ∇u\_

Answer 32

- focus on 2D, w=0 - incompressibility still holds so can introduce the streamfunction - vorticity equation reduces to Dq\_/Dt=0 - consider contours of q at rest, in the perturbed state, these straight lines become sinusoidal waves - at peaks, y\>0, ζ=-βy \< 0 =\> clockwise rotation - at troughs, y\<0, ζ=-βy \> 0 =\> anticlockwise rotation - the contour is advected as a translation to the west

Answer 33

- start with the reduced vorticity equation, Dq\_/Dt=0 - sub in for absolute vorticity q\_ = fo + βy + ∇²Ψ, this gives the non-linear Rossby-wave equation - linearise about rest state ∂/∂t(∇²Ψ) + β ∂Ψ∂/x = 0 -try 2D wave ansatz ω = -βk / [k² + l²]

Answer 34

- consider case where l=0, then ω = -β/k - phase velocity, cp = ω/k = -β/k² - group velocity, cg = ∂ω/∂k = β/k² cp ≠ cg =\> dispersive waves, but speeds are equal, cp\<0 always and cg\>0 always