Dynamics II: lecture 5

Lecture: May 17. (Monday), 14:00 Prof. Dr. Gerrit Lohmann

Tutorial: May 17. (Monday), ca. 15:30 Justus Contzen, Lars Ackermann

Time required for Sheet 2: 8 h

May 17, 14:00: Lecture 5 (online G. Lohmann, 50 min)

klick here for the zoom session

5th lecture (link)

Read Chapter 11 (The thermohaline circulation of the ocean) in Marchal and Plumb

Reading/learning might take 90 min.

Fifth lecture (link to pdf will become available here)

Content in the script: Deep ocean circulation, Conceptual models

Friction: transfer of momentum from atmosphere to oceanic Ekman layer

Vorticity dynamics for the ocean and include the wind stress term

$$D_t u - f v = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \frac{1}{\rho} \partial_z \tau_{xz}$$ $$D_t v + f u = - \frac{1}{\rho} \frac{\partial p}{\partial y} + \frac{1}{\rho} \partial_z \tau_{yz}$$

$$\frac{D}{Dt} \left( {\zeta+f}\right) - \frac{\left(\zeta+f \right)}{h} \frac{D}{Dt} h \, = \, \frac{1}{\rho} \underbrace{\left( \frac{\partial}{\partial x} \, \partial_z \tau_{yz} - \frac{\partial}{\partial y}\, \partial_z \tau_{xz} \right)}_{curl \, \partial_z \tau} \quad .$$

$$\frac{D}{Dt} \left( \frac{\zeta+f}{h}\right) = \frac{1}{\rho \, h} \, \mbox{curl} \, \partial_z \tau \,$$

$w_E$ as the Ekman vertical velocity the bottom of the Ekman layer $$w_E = - \int_{-E}^0 \frac{\partial w}{\partial z} dz = \frac{\partial}{\partial x} U_E + \frac{\partial}{\partial y} V_E$$

$\operatorname{curl} \mathbf{\tau}$ produces a divergence of the Ekman transports leading to $w_E$ at the bottom

$$w_E = \, \frac{\partial }{\partial x} \left( \frac{ \tau_{y}}{\rho \;f }\, \right) - \frac{\partial }{\partial y}\, \left( \frac{ \tau_{x}}{\rho \;f }\, \right) =\operatorname{curl}\left(\frac{\mathbf{\tau}}{\rho\;f}\right) \simeq \frac{1}{\rho\;f} \, \operatorname{curl} \mathbf{\tau}$$

The order of magnitude of the Ekman vertical velocity:

typical wind stress variation of $0.2 N m^{-2}$ per 2000 km in y-direction:

$$w_E \simeq - \frac{ \Delta \tau_{x}}{\rho \;f_0 \Delta y}\, \simeq \frac{1 }{10^3 kg m^{-3}} \frac{0.2 N m^{-2} }{10^{-4} s^{-1}\, \, 2 \cdot 10^6 m} \simeq 32 \, \, \frac{m}{yr}$$

The the conveyor is driven by deepwater formation in the northern North Atlantic.

The conveyor belt metaphor necessarily simplifies the ocean system, it is of course not a full description of the deep ocean circulation.

Broecker's concept provides a successful approach for global ocean circulation, although several features can be wrong like the missing Antarctic bottom water, the upwelling areas etc..

metaphor inspired new ideas of halting or reversing the ocean circulation and put it into a global climate context.

interpretation of Greenland ice core records indicating different climate states with different ocean modes of operation (like on and off states of a mechanical maschine).

Modelled meridional overturning streamfunction in Sv 10⁶ = m³ /s in the Atlantic Ocean. Grey areas represent zonally integrated smoothed bathymetry

It is observed that water sinks in to the deep ocean in polar regions of the Atlantic basin at a rate of 15 Sv. (Atlantic basin: 80,000,000 km² area * 4 km depth.)

– How long would it take to 'fill up' the Atlantic basin?

– Supposing that the local sinking is balanced by large-scale upwelling, estimate the strength of this upwelling.

Hint: Upwelling = area * w

– Compare this number with that of the Ekman pumping!

Timescale T to 'fill up' the Atlantic basin:

$$T = \frac{ 80 \cdot 10^{12} \, m^2 \cdot 4000 \, m}{15 \cdot 10^6 \, m^3 s^{-1}} = 2.13 \cdot 10^{10} s = 676 \;years$$

Overturning is balanced by large-scale upwelling:

$$area \cdot w = 15 \cdot 10^6 \, m^3 s^{-1}$$

$$w = 0.1875 \cdot 10^{-6} m\;s^{-1} = 5.9 \cdot 10^{-15} m \; y^{-1}.$$

Ekman pumping $$w_E \simeq 32 \, \, m \; y^{-1}.$$

$$\frac{\partial}{\partial t} v \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial y} \quad - \quad f u \quad - \quad \kappa v$$

$$\frac{\partial}{\partial t} w \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial z} \quad - \quad \frac{g}{\rho_0} (\rho -\rho_0) \quad - \quad \kappa w$$ $\kappa$ as parameter for Rayleigh friction.

$$\frac{\partial}{\partial t} v \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial y} \quad - \quad f u \quad - \quad \kappa v$$

$$\frac{\partial}{\partial t} w \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial z} \quad - \quad \frac{g}{\rho_0} (\rho -\rho_0) \quad - \quad \kappa w$$ $\kappa$ as parameter for Rayleigh friction.

Using the continuity equation
$$0 \quad = \quad \frac{\partial v}{\partial y} \quad + \quad \frac{\partial w}{\partial z}$$

$$\mbox{ one can introduce a streamfunction } \Phi (y,z): v= \partial_z \Phi; w = - \partial_y \Phi$$

The associated vorticity equation in the (y,z)-plane is therefore

$$\frac{\partial}{\partial t} \nabla^2 \Phi \, = \, - f \frac{\partial u}{\partial z} \quad + \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y} \quad - \quad \kappa \nabla^2 \Phi$$

$${\frac{\partial}{\partial t} \nabla^2 \Phi} \, = \, \underbrace{- f \frac{\partial u}{\partial z}}_{2.} \quad + \quad { \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}} \quad - \quad \underbrace{ \kappa \nabla^2 \Phi }_{4.}$$

Term 2. is absorbed into the viscous term (4.)

$$\Phi (y,z,t) \, = \, \sum_{k=1}^\infty \sum_{l=1}^\infty \Phi_{max}^{k,l} (t) \, \, \sin (\pi k y/L) \, \times \, \sin (\pi l z/H)$$ yielding a first order differential equation in time for $\Phi_{max}^{k,l} (t)$.

Simple ansatz satisfying that the normal velocity at the boundary vanishes $$\Phi (y,z,t) \, = \, \Phi_{max} (t) \, \, \sin \left(\frac{\pi y}{L}\right) \, \times \, \sin \left(\frac{\pi z}{H}\right)$$

L and H dentote the meridional and depth extend: y from 0 to L, z from 0 to H

$$\underbrace{\int_0^L dy \int_0^H dz \quad \frac{\partial}{\partial t} \nabla^2 \Phi}_{1.} \, = \, \underbrace{ \int_0^L dy \int_0^H dz \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}}_{2.} \quad - \underbrace{ \int_0^L dy \int_0^H dz \quad \kappa \nabla^2 \Phi }_{3.}$$

1. $$\frac{d}{dt} \Phi_{max} \left(\frac{\pi^2}{L^2} + \frac{\pi^2}{H^2}\right) \int\limits_0^L dy \sin \left(\frac{\pi y}{L}\right) \int\limits_0^H dz \sin \left(\frac{\pi z}{H}\right) = 4 LH \left(\frac{1}{L^2} + \frac{1}{H^2}\right) \frac{d}{dt} \Phi_{max}$$

$$\underbrace{\int_0^L dy \int_0^H dz \quad \frac{\partial}{\partial t} \nabla^2 \Phi}_{1.} \, = \, \underbrace{ \int_0^L dy \int_0^H dz \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}}_{2.} \quad - \underbrace{ \int_0^L dy \int_0^H dz \quad \kappa \nabla^2 \Phi }_{3.}$$

1. $$\frac{d}{dt} \Phi_{max} \left(\frac{\pi^2}{L^2} + \frac{\pi^2}{H^2}\right) \int\limits_0^L dy \sin \left(\frac{\pi y}{L}\right) \int\limits_0^H dz \sin \left(\frac{\pi z}{H}\right) = 4 LH \left(\frac{1}{L^2} + \frac{1}{H^2}\right) \frac{d}{dt} \Phi_{max}$$

2. $$\int\limits_0^L dy \int\limits_0^H dz \frac{g}{\rho_0} \frac{\partial \rho}{\partial y} = \frac{g}{\rho_0} \, H \, (\rho_{north} - \rho_{south}) \quad \mbox{ with } \rho_{north} = \rho (y=L) , \rho_{south} = \rho (y=0)$$

$$\frac{d}{dt} \Phi_{max} \, = \, \frac{a}{\rho_0} (\rho_{north} - \rho_{south}) \, \, - \, \, \kappa \Phi_{max}$$ with $$a = g L H^2/4(L^2 + H^2) \,$$

This shows that the overturning circulation depends on the density differences on the right and left boxes.

It is simplified to a diagnostic relation

$$\Phi_{max} = \frac{a}{\rho_0 \, \kappa} \, \, (\rho_{north} - \rho_{south}) \quad$$

because the adjustment of $\Phi_{max}$ is quasi-instantaneous due to adjustment processes, e.g. Kelvin waves.