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Changeset 11679 for NEMO/trunk/doc – NEMO

Changeset 11679 for NEMO/trunk/doc


Ignore:
Timestamp:
2019-10-11T00:15:41+02:00 (5 years ago)
Author:
agn
Message:

Wednesday's additions

File:
1 edited

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  • NEMO/trunk/doc/latex/NEMO/subfiles/chap_ZDF.tex

    r11678 r11679  
    586586\begin{figure}[!t] 
    587587  \begin{center} 
    588     \includegraphics[width=\textwidth]{Fig_ZDF_OSM_structure_of_OSBL} 
     588    \includegraphics[width=0.7\textwidth]{Fig_ZDF_OSM_structure_of_OSBL} 
    589589    \caption{ 
    590590      \protect\label{fig: OSBL_structure} 
     
    661661 
    662662\begin{equation} \label{eq:dhdt-unstable} 
    663 \frac{\partial h_\mathrm{bl}}{\partial t} + \mathbf{U}_b\nabla h_\mathrm{bl}= W_b - \frac{{\overline{w^\prime b^\prime}}_\mathrm{ent}}{\Delta B_\mathrm{bl}} 
    664 \end{equation} 
    665 where $h_\mathrm{bl}$ is the horizontally-varying depth of the OSBL, $\mathbf{U}_b$ and $W_b$ are the mean horizontal and vertical velocities at the base of the OSBL, ${\overline{w^\prime b^\prime}}_\mathrm{ent}$ is the buoyancy flux due to entrainment and $\Delta B_\mathrm{bl}$ is the difference between the buoyancy, averaged over the depth of the OSBL, and the buoyancy just below the base of the OSBL. This equation is similar to that used in mixed-layer models \cite[e.g.][]{kraus+turner67}, in which the thickness of the pycnocline is taken to be zero. \cite{grant+etal18} show that this equation for $\partial h_\mathrm{bl}/\partial t$ can be obtained from the potential energy budget of the OSBL when the pycnocline has a finite thickness. Equation \ref{eq:dhdt-unstable} is the leading term in the parametrization.%The full equation obtained by \cite{grant+etal18} includes additional terms that depend on the thickness of the pycnocline, and increase the rate of deepening of the entraining OSBL by less than $\sim20$\%.  
    666  
    667 The entrainment rate for the combination of convective and Langmuir turbulence is given by ,  
    668  
     663%\frac{\partial h_\mathrm{bl}}{\partial t} + \mathbf{U}_b\cdot\nabla h_\mathrm{bl}= W_b - \frac{{\overline{w^\prime b^\prime}}_\mathrm{ent}}{\Delta B_\mathrm{bl}} 
     664\frac{\partial h_\mathrm{bl}}{\partial t} = W_b - \frac{{\overline{w^\prime b^\prime}}_\mathrm{ent}}{\Delta B_\mathrm{bl}} 
     665\end{equation} 
     666where $h_\mathrm{bl}$ is the horizontally-varying depth of the OSBL, $\mathbf{U}_b$ and $W_b$ are the mean horizontal and vertical velocities at the base of the OSBL, ${\overline{w^\prime b^\prime}}_\mathrm{ent}$ is the buoyancy flux due to entrainment and $\Delta B_\mathrm{bl}$ is the difference between the buoyancy averaged over the depth of the OSBL (i.e.\ including the ML and pycnocline) and the buoyancy just below the base of the OSBL. This equation for the case when the pycnocline has a finite thickness, based on the potential energy budget of the OSBL, is the leading term \citep{grant+etal18} of a generalization of that used in mixed-layer models \citet[e.g.][]{kraus.turner_tellus67}, in which the thickness of the pycnocline is taken to be zero. 
     667 
     668The entrainment flux for the combination of convective and Langmuir turbulence is given by 
     669\begin{equation} \label{eq:entrain-flux} 
     670  {\overline{w^\prime b^\prime}}_\mathrm{ent} = -\alpha_{\mathrm{B}} {\overline{w^\prime b^\prime}}_0 - \alpha_{\mathrm{S}} \frac{u_*^3}{h_{\mathrm{ml}}} 
     671  + G\left(\delta/h_{\mathrm{ml}} \right)\left[\alpha_{\mathrm{S}}e^{-1.5\, \mathrm{La}_t}-\alpha_{\mathrm{L}} \frac{w_{\mathrm{*L}}^3}{h_{\mathrm{ml}}}\right] 
     672\end{equation} 
     673where the factor $G\equiv 1 - \exp (-25\delta/h_{\mathrm{bl}})(1-4\delta/h_{\mathrm{bl}})$ takes care of the lesser efficiency of Langmuir mixing when the mboundary-layer depth is much greater than the Stokes depth, and $\alpha_{\mathrm{B}}$, $\alpha_{S}$  and $\alpha_{\mathrm{L}}$ depend on the ratio of the appropriate eddy turnover time to the inertial timescale $f^{-1}$. Results from the LES suggest $\alpha_{\mathrm{B}}=0.18 F(fh_{\mathrm{bl}}/w_{*C})$, $\alpha_{S}=0.15 F(fh_{\mathrm{bl}}/u_*}$  and $\alpha_{\mathrm{L}}=0.035 F(fh_{\mathrm{bl}}/u_{*L})$, where $F(x)\equiv\tanh(x^{-1}))^{0.69}$. 
    669674 
    670675 
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