Analysis of the dynamic mechanisms of upwelling in deep ocean water caused by typhoons

Abstract

A typhoon (hurricane) is a very strong local disturbance that can affect ocean water as deep as 1000 m. According to observations and numerical simulations, the decrease in ocean temperature is believed to be caused by inertial pumping, entrainment, upwelling, and a pressure-gradient force; however, these assumptions are mostly based on modeling and do not have a clear dynamic mechanism. Therefore, in this study, the dynamic mechanism of ocean water in a highly-idealized ocean and wind field was calculated. Within the maximum wind range, downwelling in the ocean surface is caused as per the classical Ekman layer transport. Then, a pressure-gradient force drives a cyclo-geostrophic current. Because it provides centripetal force, the pressure-gradient force decreases with increasing depth. These different currents generate friction that is similar to bottom Ekman pumping, leading to upwelling. In the actual ocean, the pressure-gradient force is not only balanced by centripetal force but also baroclinic force. When these forces are balanced, the pressure-gradient force disappears, and the dynamic motion of the ocean water stops. Thus, upwelling no longer occurs. The depth at which this balance occurs is considered to be the maximum depth that a typhoon can impact. Therefore, the pressure-gradient force, caused by typhoons and then offset by centripetal and baroclinic forces, is the original dynamic of upwelling in deep ocean water.

Appendix

The wind-stress ${\varvec{\uptau}}$ is anticlockwise associated with the wind field of the typhoon and we assume that ${\varvec{\uptau}}$ is axis symmetry and divergence free, i.e. $\nabla \cdot {\varvec{\uptau}}$=0, as shown in Fig. 11.

Fig. 11

The idealized wind-stress and coordinate system translation

The Cartesian coordinate system ($x$, $y$, $z$) can be translated into a cylinder coordinate system ($r$, $\theta$, $z$).

$$\left\{\begin{array}{c}r=\sqrt{{x}^{2}+{y}^{2}}\\ \theta =arctan\frac{y}{x}\\ z=z\end{array}\right.$$

(23)

The magnitude of ${\varvec{\uptau}}$ is $\tau$ and the direction is azimuthal, then:

$$\left\{\begin{array}{l}{\tau }_{x}=-\tau \;sin\theta \\ {\tau }_{y}=\tau \;cos\theta \end{array},\right.$$

(24)

and

$$\left\{\begin{array}{c}x=r\;cos\theta \\ y=r\;sin\theta \end{array}.\right.$$

(25)

Therefore,

$$\frac{\partial r}{\partial x}=\frac{\partial \sqrt{{x}^{2}+{y}^{2}}}{\partial x}=\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}=\frac{x}{r}=cos\theta$$

(26)

$$\frac{\partial r}{\partial y}=\frac{\partial \sqrt{{x}^{2}+{y}^{2}}}{\partial y}=\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}=\frac{y}{r}=sin\theta ,$$

(27)

$$\frac{\partial \theta }{\partial x}=\frac{\partial \left(arctan\frac{y}{x}\right)}{\partial x}=-\frac{sin\theta }{r},$$

(28)

$$\frac{\partial \theta }{\partial y}=\frac{\partial \left(arctan\frac{y}{x}\right)}{\partial y}=\frac{cos\theta }{r},$$

(29)

$$\frac{\partial \theta }{\partial r}=0,$$

(30)

$$\frac{\partial \tau }{\partial \theta }=0.$$

(31)

According to the derivation chain rule, we can obtain:

$$\frac{\partial {\tau }_{y}}{\partial x}=\frac{\partial {\tau }_{y}}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial {\tau }_{y}}{\partial \theta }\frac{\partial \theta }{\partial x}=\frac{\partial \tau }{\partial r}{cos}^{2}\theta +\frac{\tau }{r}{sin}^{2}\theta ,$$

(32)

$$-\frac{\partial {\tau }_{x}}{\partial y}=\frac{\partial {\tau }_{x}}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial {\tau }_{x}}{\partial \theta }\frac{\partial \theta }{\partial y}=\frac{\partial \tau }{\partial r}{sin}^{2}\theta +\frac{\tau }{r}{cos}^{2}\theta ,$$

(33)

$$\frac{\partial {\tau }_{x}}{\partial x}=\frac{\partial {\tau }_{x}}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial {\tau }_{x}}{\partial \theta }\frac{\partial \theta }{\partial x}=-\frac{\partial \tau }{\partial r}cos\theta sin\theta +\frac{\tau }{r}sin\theta cos\theta$$

(34)

$$\frac{\partial {\tau }_{y}}{\partial y}=\frac{\partial {\tau }_{y}}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial {\tau }_{y}}{\partial \theta }\frac{\partial \theta }{\partial y}=\frac{\partial \tau }{\partial r}sin\theta cos\theta -\frac{\tau }{r}cos\theta sin\theta .$$

(35)

Thus, the differential formulae of the surface vertical velocity is:

$$\begin{array}{c}\frac{\partial {w}_{s}}{\partial z}=-\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\\ =-\frac{\sqrt{2}}{{\rho }_{0}fd}{e}^\frac{z}{d}\left[\frac{\partial {\tau }_{x}}{\partial x}\mathrm{sin}\left(\frac{\pi }{4}+\frac{z}{d}\right)+\frac{\partial {\tau }_{y}}{\partial x}\mathrm{cos}\left(\frac{\pi }{4}+\frac{z}{d}\right)-\frac{\partial {\tau }_{x}}{\partial y}\mathrm{cos}\left(\frac{\pi }{4}+\frac{z}{d}\right)+\frac{\partial {\tau }_{y}}{\partial y}\mathrm{sin}\left(\frac{\pi }{4}+\frac{z}{d}\right)\right]\\ =-\frac{\sqrt{2}}{{\rho }_{0}fd}{e}^\frac{z}{d}\left[\left(\frac{\partial {\tau }_{y}}{\partial x}-\frac{\partial {\tau }_{\mathrm{x}}}{\partial y}\right)\mathrm{cos}\left(\frac{\pi }{4}+\frac{z}{d}\right)+\left(\frac{\partial {\tau }_{x}}{\partial x}+\frac{\partial {\tau }_{y}}{\partial y}\right)\mathrm{sin}\left(\frac{\pi }{4}+\frac{z}{d}\right)\right]\\ =-\frac{\sqrt{2}}{{\rho }_{0}fd}{e}^\frac{z}{d}\left[\begin{array}{c}\left(\frac{\partial \tau }{\partial r}{\mathrm{cos}}^{2}\theta +\frac{\tau }{r}{\mathrm{sin}}^{2}\theta +\frac{\partial \tau }{\partial r}{\mathrm{sin}}^{2}\theta +\frac{\tau }{r}{\mathrm{cos}}^{2}\theta \right)\mathrm{cos}\left(\frac{\pi }{4}+\frac{z}{d}\right)+\\ \left(-\frac{\partial \tau }{\partial r}\mathrm{cos}\theta \mathrm{sin}\theta +\frac{\tau }{r}\mathrm{sin}\theta \mathrm{cos}\theta +\frac{\partial \tau }{\partial r}\mathrm{sin}\theta \mathrm{cos}\theta -\frac{\tau }{r}\mathrm{cos}\theta \mathrm{sin}\theta \right)\mathrm{sin}\left(\frac{\pi }{4}+\frac{z}{d}\right)\end{array}\right]\\ =-\frac{\sqrt{2}}{{\rho }_{0}fd}{e}^\frac{z}{d}\left(\frac{\partial \tau }{\partial r}+\frac{\tau }{r}\right)cos\left(\frac{\pi }{4}+\frac{z}{d}\right)\end{array}.$$

(36)

If it is similar to the surface, we can obtain the differential formulae of the bottom Ekman layer using the following:

$$\begin{array}{c}\frac{\partial {w}_{b}}{\partial z}=-\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\\ =-\frac{\partial \overline{u}}{\partial x }\left(1-{e}^{-\frac{z}{d}}cos\frac{z}{d}\right)+\frac{\partial \overline{v}}{\partial x}{e }^{-\frac{z}{d}}sin\frac{z}{d}-\frac{\partial \overline{v}}{\partial y }\left(1-{e}^{-\frac{z}{d}}cos\frac{z}{d}\right)-\frac{\partial \overline{u}}{\partial y}{e }^{-\frac{z}{d}}sin\frac{z}{d}\\ ={e}^{-\frac{z}{d}}\left(\frac{\partial {v}_{\theta }}{\partial r}+\frac{{v}_{\theta }}{r}\right)sin\frac{z}{d}\end{array}.$$

(37)

The solution of Eq. (14) can be calculated as follows:

$$\begin{array}{c}{w}_{s}=\int \frac{\partial {w}_{s}}{\partial z}dz\\ =-\frac{\sqrt{2}}{{\rho }_{0}fd}\left(\frac{\partial \tau }{\partial r}+\frac{\tau }{r}\right)\int {e}^\frac{z}{d}\mathrm{cos}\left(\frac{\pi }{4}+\frac{z}{d}\right)dz\\ \begin{array}{c}=-\frac{\sqrt{2}}{{\rho }_{0}fd}\left(\frac{\partial \tau }{\partial r}+\frac{\tau }{r}\right)\left({e}^{-\frac{\pi }{r}}d\int {e}^{\frac{z}{d}+\frac{\pi }{4}}\mathrm{cos}\left(\frac{z}{d}+\frac{\pi }{4}\right)d\left(\frac{z}{d}+\frac{\pi }{4}\right)\right)\\ =-\frac{\sqrt{2}}{{\rho }_{0}fd}\left(\frac{\partial \tau }{\partial r}+\frac{\tau }{r}\right){e}^{-\frac{\pi }{r}}d\left(\frac{1}{2}{e}^{\frac{z}{d}+\frac{\pi }{r}}\left(\mathrm{sin}\left(\frac{z}{d}+\frac{\pi }{r}\right)+\mathrm{cos}\left(\frac{z}{d}+\frac{\pi }{4}\right)\right)+{C}_{0}\right)\\ =-\frac{1}{{p}_{0}f}{e}^\frac{z}{d}\left(\frac{\partial \tau }{\partial r}+\frac{\tau }{r}\right)\mathrm{cos}\frac{z}{d}+{C}_{1}\end{array}\end{array}$$

(38)

The solution of Eq. (15) can be calculated as follows:

$$\begin{array}{c}{w}_{b}=\int \frac{\partial {w}_{b}}{\partial z}dz\\ =\int \left(\frac{\partial {v}_{\theta }}{\partial r}+\frac{{v}_{\theta }}{r}\right){de}^{-\frac{z}{d}}sin\left(-\frac{z}{d}\right)d\left(-\frac{z}{d}\right)\\ =\left(\frac{\partial {v}_{\theta }}{\partial r}+\frac{{v}_{\theta }}{r}\right)\left(\frac{d}{2}{e}^{-\frac{z}{d}}\left(sin\left(-\frac{z}{d}\right)-cos\left(-\frac{z}{d}\right)\right)+C\right)\\ =\left(\frac{\partial {v}_{\theta }}{\partial r}+\frac{{v}_{\theta }}{r}\right)\left(-\frac{\sqrt{2}d}{2}{e}^{-\frac{z}{d}}sin\left(\frac{z}{d}+\frac{\pi }{4}\right)+C\right)\end{array}$$

(39)

In the cylinder coordinate system, the $W$ in the entire water column is obtained as follow:

$$\frac{\partial p}{\partial x}=\frac{\partial p}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial p}{\partial \theta }\frac{\partial \theta }{\partial x}$$

(40)

$$\frac{\partial p}{\partial y}=\frac{\partial p}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial p}{\partial \theta }\frac{\partial \theta }{\partial y}$$

(41)

Using Eqs. (4)-(7), the equations are now

$$\frac{\partial p}{\partial x}=cos\theta \frac{\partial p}{\partial r}-\frac{sin\theta }{r}\frac{\partial p}{\partial \theta }$$

(42)

$$\frac{\partial p}{\partial y}=sin\theta \frac{\partial p}{\partial r}+\frac{cos\theta }{r}\frac{\partial p}{\partial \theta }.$$

(43)

Then, we obtain

$$\frac{{\partial }^{2}p}{\partial {x}^{2}}=\frac{\partial }{\partial x}\left(\frac{\partial p}{\partial x}\right)=cos\theta \frac{\partial }{\partial r}\left(\frac{\partial p}{\partial r}cos\theta -\frac{sin\theta }{r}\frac{\partial p}{\partial \theta }\right)-\frac{sin\theta }{r}\frac{\partial }{\partial \theta }\left(\frac{\partial p}{\partial r}cos\theta -\frac{sin\theta }{r}\frac{\partial p}{\partial \theta }\right)$$

(44)

$$\frac{{\partial }^{2}p}{\partial {y}^{2}}=\frac{\partial }{\partial y}\left(\frac{\partial p}{\partial y}\right)=sin\theta \frac{\partial }{\partial r}\left(sin\theta \frac{\partial p}{\partial r}+\frac{cos\theta }{r}\frac{\partial p}{\partial \theta }\right)+\frac{cos\theta }{r}\frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial p}{\partial r}+\frac{cos\theta }{r}\frac{\partial p}{\partial \theta }\right)$$

(45)

The four terms on the right side of the equations (B-14) and (B-15) are as follows:

$$cos\theta \frac{\partial }{\partial r}\left(\frac{\partial p}{\partial r}cos\theta -\frac{sin\theta }{r}\frac{\partial p}{\partial \theta }\right)={cos}^{2}\theta \frac{{\partial }^{2}p}{\partial {r}^{2}}+\frac{sin\theta cos\theta }{{r}^{2}}\frac{\partial p}{\partial \theta }-\frac{sin\theta cos\theta }{r}\frac{{\partial }^{2}p}{\partial r\partial \theta }$$

(46)

$$-\frac{sin\theta }{r}\frac{\partial }{\partial \theta }\left(\frac{\partial p}{\partial r}cos\theta -\frac{sin\theta }{r}\frac{\partial p}{\partial \theta }\right)=-\frac{sin\theta cos\theta }{r}\frac{{\partial }^{2}p}{\partial r\partial \theta }+\frac{{sin}^{2}\theta }{r}\frac{\partial p}{\partial r}+\frac{sin\theta cos\theta }{{r}^{2}}\frac{\partial p}{\partial \theta }+\frac{{sin}^{2}\theta }{{r}^{2}}\frac{{\partial }^{2}p}{\partial {\theta }^{2}}$$

(47)

$$sin\theta \frac{\partial }{\partial r}\left(sin\theta \frac{\partial p}{\partial r}+\frac{cos\theta }{r}\frac{\partial p}{\partial \theta }\right)={sin}^{2}\theta \frac{{\partial }^{2}p}{\partial {r}^{2}}-\frac{sin\theta cos\theta }{{r}^{2}}\frac{\partial p}{\partial \theta }+\frac{sin\theta cos\theta }{r}\frac{{\partial }^{2}p}{\partial r\partial \theta }$$

(48)

$$\frac{cos\theta }{r}\frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial p}{\partial r}+\frac{cos\theta }{r}\frac{\partial p}{\partial \theta }\right)=\frac{{cos}^{2}\theta }{r}\frac{\partial p}{\partial r}+\frac{sin\theta cos\theta }{r}\frac{{\partial }^{2}p}{\partial \theta \partial r}-\frac{sin\theta cos\theta }{{r}^{2}}\frac{\partial p}{\partial \theta }+\frac{{cos}^{2}\theta }{{r}^{2}}\frac{{\partial }^{2}p}{\partial {\theta }^{2}}.$$

(49)

Therefore, we get

$${\nabla }^{2}p=\frac{{\partial }^{2}p}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial p}{\partial r}+\frac{1}{{r}^{2}}\frac{{\partial }^{2}p}{\partial {\theta }^{2}}$$

(50)

Along an isobar, the pressure is const, Thus

$$\frac{\partial p}{\partial \theta }=0$$

(51)

and

$$\frac{{\partial }^{2}p}{\partial {\theta }^{2}}=0.$$

(52)

Then, we obtain the Laplace operator of $p$ in the cylinder coordinate system:

$${\nabla }^{2}p=\frac{{\partial }^{2}p}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial p}{\partial r}.$$

(53)

So, we get the vertical velocity caused by bottom Ekman at the role of $p$ as follow:

$$W=\frac{{d}_{B}}{2{\rho }_{0}f}\left(\frac{{\partial }^{2}p}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial p}{\partial r}\right).$$

(54)