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Next: 4.3 Growth rates Up: 4. Discussion of Solutions: Previous: 4.1 Frequency and scales

Subsections

   
4.2 Coupling constants

4.2.1 Mechanical

Let $\alpha ^{\prime }$ scale the stress of the wind, $\tau $, to the surface wind speed us thus:

 \begin{displaymath}\tau ^{\prime }=\alpha ^{\prime }{u_{s}}^{\prime }
\end{displaymath} (55)

To deduce a value for $\alpha ^{\prime }$, consider the bulk aerodynamic drag law for the total (mean + anomaly) windstress (see Gill, 1982):

\begin{displaymath}\overline{\tau }+\tau ^{\prime }=c_{_{D}}\rho _{a}(\overline{u_{s}}+{u_{s}}
^{\prime })^{2}
\end{displaymath}

where c<<1113>>D is the drag coefficient. After linearizing about the mean $
\overline{u_{s}}$, we obtain

 \begin{displaymath}\tau ^{\prime }=2c_{_{D}}\rho _{a}\overline{u_{s}}\cdot {u_{s}}^{\prime }
\end{displaymath} (56)

allowing us to identify:


\begin{displaymath}\alpha ^{\prime }=2c_{_{D}}\rho _{a}\overline{u_{s}}
\end{displaymath}

Comparing (17), (55) and (56), we see that:

\begin{displaymath}\alpha =\frac{\alpha ^{\prime }}{\rho _{o0}h}=\frac{2c_{_{D}}\rho _{a}
\overline{u_{s}}}{\rho _{o0}h}
\end{displaymath}

In accord with observations, for $\overline{u_{s}}=5\mbox{~m/s}$, h=500 m, we find that $\alpha \approx \frac{2\times 1.3\cdot 10^{-3}\times 1\times
5}{10^{3}\times 500}=3\cdot 10^{-8}$ $s^{-1}=\frac{1}{1.1 \mbox{~years}}$ if $c_{_{D}}=1.3\cdot10^{-3}$.

4.2.2 Thermal Equilibration

The inverse damping time-scale of a PV anomaly, $\Gamma \equiv \frac{4\gamma_a
}{\kappa ^{2}L_a^2}$, (32), depends on the scale of the anomaly relative to the deformation radius and the radiative-convective restoring time-scale. Inserting typical numbers we find

\begin{displaymath}\Gamma \equiv \left[
4\frac{1}{14\mbox{\scriptsize ~days}}\le...
...ot10^{-6}\mbox{~s}^{-1}=\frac{1}{1.5 \mbox{\scriptsize ~days}}
\end{displaymath}

This time-scale becomes shorter the greater the scale of the anomaly relative to the deformation radius.

4.2.3 SST

By putting numbers into (38) we find that the SST coupling parameter $r \equiv H/(2\varepsilon \theta_{a0}h) \approx
10^{4}\left(2\cdot10^{-4}\cdot 290 \cdot 500\right)^{-1} \approx 340$. A reasonable value for a is $a \equiv
(gh\varepsilon/f)\frac{\partial}{\partial y}\overline{\mbox{SST}} =
10\cdot500\cdot10^{-4}/10^{-4})\cdot3\cdot10^{-6} = 0.015 \mbox{~m~s}^{-1}$. With $k = 5\cdot10^{-7}\mbox{~m}^{-1}$, the advection timescale is $a k =
7.5\cdot10^{-9}\mbox{~s}^{-1}$. In section 2.4.2, we established the entrainment parameter $\gamma_e = 1.3\cdot10^{-7} \mbox{s}^{-1}$ and the air-sea flux parameter $\gamma_o = 5\cdot10^{-8} \mbox{s}^{-1}$. If $\sigma \sim \omega_r
\sim 2\cdot10^{-8}\mbox{~s}^{-1}$, then for this choice of parameters $\gamma_e \gg
\sigma$, $\gamma_e \gg ak$, $\gamma_e > \gamma_o$, so the entrainment solution should dominate in the full dispersion relation (40), perhaps with some contribution from air-sea flux. Furthermore, the second and third terms beneath the radical in (41) are smaller than the first, so the approximation leading to (47) and (48) should be valid. We now compute growth rates as a function of wavelength and other parameters to see if this is indeed the case.


next up previous
Next: 4.3 Growth rates Up: 4. Discussion of Solutions: Previous: 4.1 Frequency and scales
Jason C Goodman
1998-03-09