Book: Dynamical Oceanography Author: Dijkstra, H. A.; email: [email protected] Chapter: 5, Exercise: 5.2: An alternative approach to the bottom Ekman layer Version: 2 Consider an ocean with water of constant density ρ above a flat bottom (at z∗ = 0). The horizontal velocities outside the Ekman layer (where the flow is hydrostatic and geostrophic) are indicated by (ug∗; vg∗). The total velocity can be written as u∗ = ug∗ + uE∗; v∗ = vg∗ + vE∗ where uE∗ en vE∗ are the horizontal Ekman velocities. a. Show that the Ekman velocities can be determined from @2u −f v = A E∗ 0 E∗ V @z2 @2v f u = A E∗ 0 E∗ V @z2 If we neglect inertia and horizontal friction, the momentum equations become 1 −fv = − p + A u ρ x V zz 1 fu = − p + A v ρ y V zz 2 2 where px = @p=@x, uzz = @ u=@z , etc. In the interior we have the geostrophic balance, given by: 1 −fv = −fv = − p g ρ x 1 fu = fu = − p g ρ y The geostrophic flow ug, vg and the horizontal gradients are independent of z, because: 1 g −f(v ) = − p = (ρ) = 0 g z ρ xz ρ x 1 g f(u ) = − p = (ρ) = 0 g z ρ yz ρ y due to the assumption of homogeneous density. Because the momentum equations are linear in the velocity we can superpose the velocity field in a geostrophic and an Ekman part: u = ug + uE; v = vg + vE If we subtract the geostrophic balance from the momentum balance, we end up with: −fvE = AV (uE + ug)zz = AV (uE)zz fuE = AV (vE + vg)zz = AV (vE)zz So, the Ekman velocity arises from a balance between the Coriolis force and the frictional force. ~ x y b. Determine the Ekman transport ME∗ = (ME∗;ME∗) through vertical integration and show that the direction of this transport is perpendicular to the bottom shear stress. 1 Cross differentiation the result under a. gives one equation for uE: 2 2 f uE + AV (uE)zzzz = 0 Solutions of the form λz uE = u0e lead to 2 2 4 f + AV λ = 0 so that we find for λ: 1 λ = ±(1 ± i) δE in which the Ekman depth δE is given by: s 2A δ = v E jfj Because we do not allow exponentially growing solutions with z, (the influence of friction vanishes in the interior) two solutions are unphysical. The other two solutions are fixed by the condition that the velocity should vanish at the boundary, so uE = −ug at the boundary. If we assign a coordinate system to the flow that originates at the bottom the following solution is found: z f z −z/δE u = ug − e (ug cos + vg sin ) δE jfj δE z f z −z/δE v = vg − e (vg cos − ug sin ) δE jfj δE Close to the boundary the solution is: f z u = (ug − vg) jfj δE f z v = ( ug + vg) jfj δE so the solution is at 450 of the interior geostrophic flow, rotated to the left. (That is, for the northern hemisphere. For the souther hemisphere, the rotation is to the right). Going further up in the water column both velocity components increase, to reach a maximum at z = 3πδE=4. This maximum is in fact larger than the geostrophic interior flow. From then on the velocity decreases again to the interior geostrophic velocity. The Ekman transport in the Ekman layer can be found by integrating the Ekman velocity over this layer: Z 1 x δE f ME = (u − ug)dz = − (ug + vg) 0 2 jfj Z 1 y δE f ME = (v − vg)dz = ( ug − vg) 0 2 jfj So, the Ekman transport is proportional to the layer depth and the geostrophic interior velocity. c. Show that for vg∗ = 0 and constant ug∗, the solution of the Ekman layer velocities is given by z −z∗/δE ∗ uE∗ = −ug∗e cos δE z −z∗/δE ∗ vE∗ = ug∗e sin δE 2 q 2AV where δE = is the bottom Ekman layer thickness. f0 For vg = 0, the solution of the Ekman layer velocities is given by z −z/δE uE = −uge cos δE z −z/δE vE = uge sin δE d. Check that the total velocity field (u; v) satisfies the boundary conditions (z = 0) : u∗ = v∗ = 0 (z 1) : u∗ = ug∗; v∗ = 0 and make a sketch of the total velocity field in the Ekman layer. These can be easily checked by using u = uE + ug and v = vE + vg and using the expressions for uE and vE as obtained under c. 3.