Lesson 5 Boussinesq Approximation

Boussinesq approximation
Recall for an incompressible fluid: ∇ ⋅ v = 0
Conservation of Momentum

Derivation of the Equations of Motion
Assume density variations are small and (EOM) therefore can compare local variations of
Navier-Stokes equations the density field to uniform background
Must be able to identify terms and coordinate density ρ ′ systems << 1 ρ 0

Valid for nearly all circumstances in the ocean and lower atmosphere

Conservation of Momentum Conservation of Momentum

Newton’s Second Law
Use this to derive our equations of motion

Relates the rate of change of absolute
Navier-Stokes Equations

momentum following the motion in an
Let’s re-examine the net forces acting on a inertial reference frame to the sum of the fluid parcel forces acting on the fluid =
This information helps to construct equations p mv that calculate the net force - and therefore dv dp acceleration - acting on a parcel ∑F = ma = m =
Sum the forces acting in each direction and dt dt equate it to the acceleration of the parcel

1 Primary Forces Where are we going…

Lesson 5
Vertical Components Vector form of EOM
Gravity

Centrifugal force Spherical form of EOM
PGF

Coriolis Shallow Cartesian (rectangular)
Horizontal Componets

PGF Primitive f-plane
Viscous Equations β-plane

Coriolis

Equations of Motion: Goal EOM: Derivation

Now, to deal with the rotating frame of Du 1 ∂P = − + υ∇ 2u + (2Ωv sin φ − 2Ωwcos φ) reference.... Dt ρ ∂x

Inertial (absolute) frame Dv 1 ∂P = − + υ∇ 2v − (2Ωusin φ) = ˆ + ˆ + ˆ Dt ρ ∂y A Axi Ay j Azk
Non-inertial (rotating) frame ∂ Dw 1 P 2 = ′ˆ ′ + ′ˆ ′ + ′ˆ ′ = − + υ∇ w + (2Ωucos φ) − g A Ax i Ay j Az k Dt ρ ∂z

2 EOM: Derivation EOM: Derivation
Relationship between inertial and rotating DaU a = DU r + Ω × − Ω2 frames of reference: 2 U r r Dt Dt Da A Dr A = + Ω × A
Acceleration following the motion in an Dt Dt inertial frame is equal to the acceleration following the motion in a rotating frame
Relationship between absolute velocity plus the coriolis acceleration and and velocity relative to Earth centripetal acceleration = + Ω × U a U r r

EOM: Vector EOM: Spherical Coordinates

Rewrite to get N2L relative to a rotating
Assume the Earth’s departure from a coordinate frame: sphere is negligible

Vector EOM
Expand vector form of EOM so that the surface of the Earth corresponds to a coordinate surface
λ φ dU r 1 2 Coordinate axes ( , , z): = − Ω × − ∇ + + υ∇ −π ≤ λ ≤ π 2 U r P g U r
λ: longitude ( ) ρ π π dt φ (− ≤ φ ≤ )
: latitude 2 2 1 2 3 4 5
z: height above Earth’s surface

a: radius of the Earth

Unit vectors to describe motion

3 EOM: Spherical Coordinates EOM: Spherical Coordinates

Relative velocity vector becomes:
Zonal (East-West) distance = ˆ + ˆ + ˆ V u i v j w k dx = acos φdλ
Where:
Meridional (North-South) distance dλ u ≡ acos φ dy = ad φ dt dφ
Vertical distance v ≡ a dt dz = da dz w ≡ dt

EOM: Derivation EOM: Derivation Diˆ ∂iˆ u Dˆj ∂ˆj ∂ˆj utan φ v = u = (sin φˆj − cos φkˆ ) = u + v = − iˆ − kˆ Dt ∂x acos φ Dt ∂x ∂y a a
Determine the magnitude and direction
Determine the magnitude and direction

4 EOM: Derivation EOM: Derivation

Dkˆ ∂kˆ ∂kˆ u v
Substitute into relative velocity equation: = u + v = iˆ + ˆj Dt ∂x ∂y a a DV Du uv tan φ uw Dv u2 tan φ vw Dw u2 + v 2 = ( − + )iˆ + ( + + ) ˆj + ( − )kˆ
Determine the magnitude and direction Dt Dt a a Dt a a Dt a

Describes the spherical expansion of the acceleration following the motion
Recall, our vector EOM had our force terms…now we have to expand the force terms to get the complete EOM in spherical coordinates

EOM: Spherical Where are we going…

Du uv tan φ uw 1 ∂P − + = − + 2Ωv sin φ − 2Ωw cos φ + F Vector form of EOM Dt a a ρ ∂x x

Spherical form of EOM 2 φ ∂ Dv + u tan + vw = − 1 P − Ω φ + 2 usin Fy Dt a a ρ ∂y Shallow Cartesian (rectangular)

Lesson 6 Dw u2 + v 2 1 ∂P − = − − g + 2Ωucos φ + F Primitive f-plane Dt a ρ ∂z z Equations β-plane

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