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Karten 35 Karten 1 Lernende English Universität 21.07.2018 / 14.08.2018 Keine Angabe
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shallow water equations for uniform density and flat ocean bottom

complete set of the shallow water equations:

$$u_t-fv=-g\eta_x \\ v_t+fu=-g\eta_y \\ \eta_t+H(u_x+v_y)=0$$

derivation:

1. hydrostatic approximation: $$.\\ (\frac{H}{L})^2<<1\\$$

2. horizontal momentum equations for small perturbations about a state of rest (working with total pressure):

$$u_t-fv=-\frac{p_x}{p_0} \\ v_t +fu=-\frac{p_y}{p_0}$$

3. hydrostatic equation:

$$0=-\frac{\partial p}{\partial z}-g\rho_0 \\ =>p=p_a+g\rho_0(\eta-z) \\ =>p_x=(p_a)_x+g\rho_0\eta_x$$

it follows:

$$u_t-fv=-g\eta_x-(\frac{p_{ax}}{\rho_0}) \\ v_t-fu=-g\eta_y-(\frac{p_{ay}}{\rho_0})$$

3. the kinematic boundary condition at the sea surface:

the total change of sea surface height is äquivalent to the vertical motion w:

$$\eta_t+u\eta_x+v\eta_y=w$$

for small perturbations about a state of rest it follows:

$$\eta_t=w$$

4. integrating the continuity equation and applying $$\eta_t=0$$ yields:

$$\int_{-H}^{0}(u_x+v_y)\;dz=-\int_{-H}^{0}w_z\;dz=-\eta_t \\ => \eta_t+H(u_x+v_y)=0$$

5. assuming uniform atmospheric pressure:

$$i.e.\;\;p_{ax}=p_{ay}=0$$

What is the main concept of a layer model?

• A natural way to simulate the ocean circulation is to assume a two-layer fluid. The main thermocline is used as the interface. The lower layer is very thick and the water in the lower layer moves much slower than in the upper.
• reduced gravity model: the lower layer is in stagnation-> One active layer only! (also called the 1 1/2- layer model)
• If the model has two levels and both are in motion than its the 2 layer model...

S.262

What is the rigid-lid approximation? $$p_a$$

The upper boundary condition of the ocean model is moved from the free surface $$z=\zeta$$ to a flat surface $$z=0$$.

If we assume the lowest layer to be stagnant then this leads to a drop out of the atmospheric pressure in the horizontal pressure gradient terms.

$$p_a=p_{a,0}+\rho g \zeta$$

This leads for the 1 1/2-layer model again to the shallow water equations:

$$u_t-fv= -g'h_x \\ v_t+fu=-g'h_y \\ h_t+H_1(u_x+v_y)=0$$

g' is the reduced gravity:

$$g'=g(\rho_1-\rho_2)/\rho_0$$

Explain the meridional flow driven by Ekman pumping.

1. We find prevailing westerlies at mid-latitudes and easterlies at low and high latitudes.
2. This wind stress pattern drives poleward Ekman flows at low and high latitudes but an equatorward Ekman flow at mid-latitudes.
3. The meridional convergence of the Ekman flux in the upper ocean gives rise to the Ekman pumping (compression of the water column) and upwelling (expansion) below the base of the Ekman layer.
4. In the basin interior, relative vorticity is negligible, so the potential vorticity can be written as  $$f/h$$ .
• A compressed water column needs to move equatorward and a lengthened water column needs to move polewards to maintain potential vorticity.

=> Ekman pumping in the subtropical basin drives an equatorward flow in the ocean interior.

=> Ekman upwelling in the subpolar basin drives a poleward flow in the ocean interior.

figure in Huang et al. 2010 S.286

Including wind forcing to the shallow water equations!

$$u_t-fv=-\frac{p'_x}{\rho_0}+\frac{\partial X}{\partial z} \\ v_t+fu=-\frac{p'_y}{\rho_0}+\frac{\partial Y}{\partial z}$$

with

$$X=\nu u_z,\;\;Y=\nu v_z$$

where $$\nu$$ is a vertical eddy viscosity (but could also be something else)

The connection between vertical eddy viscosity and the surface wind stress is:

$$\nu u_z=\tau^x_s/\rho_0\;\;\;\nu v_z=\tau^y_s/\rho_0$$

This means: $$\rho_0(X,Y)$$ is the horizontally acting stress at each depth

In this course we often set:

$$\nu=\left\{ \begin{array}{c} \nu_0\;\;\;for\;-H_E\leq z\leq0\\0\;\;\; otherwise \end{array} \right.$$

The balance

$$-fv=(\nu u_z)_z\\ fu=(\nu v_z)_z$$

implies, for uniform $$\nu$$ the scaling

$$fU=\frac{\nu U}{h^2}=>h=\sqrt{\frac{\nu}{f}}$$  (h is the vertical scale)

i.e. for uniform $$\nu$$ variations in the vertical take place in a scale $$\sqrt{\frac{\nu}{f}}$$ . If this scale is much larger than the depth over which $$\nu$$ is non-zero, then the horizontal velocity will be essentially independent of depth over the region where $$\nu$$ is non-zero, in this case the depth  $$H_E$$ .

For this case we can treat wind stress as a body force over the depth $$H_E$$ . Which means the wind stress forcing is constant for each layer whithin the Ekman Layer.

$$\left(\frac{\partial X}{\partial z},\frac{\partial Y}{\partial z}\right)=\frac{G(z)}{\rho_0H_E}(\tau_s^x,\tau^y_s)$$

with

$$G(z)=\left\{ \begin{array}{c} 1\;\;-H_E\leq z\leq0\\0\;\;otherwise \end{array} \right.$$

for the 1 1/2 layer model follows with wind forcing:

$$u_t-fv=-g'h_x+\frac{\tau^x_s}{\rho_1H_1}\\ v_t+fu=-g'h_y+\frac{\tau_s^y}{\rho_1H_1}\\ h_t+H_1(u_x+v_y)=0$$

Ekman dynamics!

Consider the Ekman part of the flow:

$$u_t-fv=-\frac{p'_x}{\rho_0}+\frac{\partial X}{\partial z}\\ v_t+fu=-\frac{p'_y}{\rho_0}+\frac{\partial Y}{\partial z}$$

is going to:

$$u_t-fv=+\frac{\partial X}{\partial z}\\ v_t+fu=-\frac{\partial Y}{\partial z}$$

1. Assuming that the flow is al confined above a depth $$H_E$$ , the EKMAN TRANSPORT, ($$(U_E,V_E)$$, is defined as:

$$(U_E,V_E)=\int\limits_{-H_E}^0(u_E,v_E)dz$$

$$(u_E,v_E)$$i is the Ekman Velocity.

2. Assuming $$(X,Y)=0$$below the Ekman depth, it follows that

$$U_{Et}-fV_E=\frac{\tau^x_s}{\rho_0H_E}\\ V_{Et}+fU_E=\frac{\tau^y_s}{\rho_0H_E}$$

-> In steady state, the Ekman transport is at right angles to the surface wind stress, i.e.

$$-fV_E=\frac{\tau^x_s}{\rho_0H_E}\\ fU_E=\frac{\tau^x_s}{\rho_0H_E}$$

Geostrophic dynamics: What says the Taylor-Proudman Theorem?

The geostrophic balance is expressed by:

$$-fv_g=-\frac{1}{\rho_0}p_x\\ fu_g=-\frac{1}{\rho_0}P_y$$

where $$(u_g,v_g)$$ is the geostrophic flow. It follows that:

$$\frac{\partial u_g}{\partial x}+\frac{\partial v_g}{\partial y}=-\frac{\beta v}{f}$$

On a f-plane,  $$\beta=0$$ and hence

$$\frac{\partial u_g}{\partial x}+\frac{\partial v_g}{\partial y}=0$$

Using $$u_x+v_y+w_z=0$$ it follows for the f-plane:

$$w_z=0$$

This means that if vertical velocity is somewhere zero (as it is for the bottom the case), than it is everywhere zero! It follows that flow will tend to go around topography and not over topography!

Furthermore, with the thermal wind equations it can be shown that for a fluid with uniform density, that there is no variation in the geostrophic flow in the vertical (no geostrophic shear!) which means that above a bump, the flow will be stagnant at all depths!

The Taylor-Proudman Theorem states that flow tends to go around bumps rather than over them!

Write down the thermal wind equations!

Combine the geostrophic balance

$$-fv=-\frac{1}{\rho_0}p_x\\fu=-\frac{1}{\rho_0}p_y$$

with the hydrostatic balance $$0=-p_z-g\rho$$ to give

$$fv_z=-\frac{g}{\rho_0}\rho_x\\fu_z=\frac{g}{\rho_0}\rho_y$$