Ocean response to typhoon Nuri (2008) in western Pacific and South China Sea

Abstract

Typhoon Nuri formed on 18 August 2008 in the western North Pacific east of the Philippines and traversed northwestward over the Kuroshio in the Luzon Strait where it intensified to a category 3 typhoon. The storm weakened as it passed over South China Sea (SCS) and made landfall in Hong Kong as a category 1 typhoon on 22 August. Despite the storm’s modest strength, the change in typhoon Nuri’s intensity was unique in that it strongly depended on the upper ocean. This study examines the ocean response to typhoon Nuri using the Princeton Ocean Model. An ocean state accounting for the sea-surface temperature (SST) and mesoscale eddy field prior to Nuri was constructed by assimilating satellite SST and altimetry data 12 days before the storm. The simulation then continued without further data assimilation, so that the ocean response to the strong wind can be used to understand processes. It is found that the SST cooling was biased to the right of the storm’s track due to inertial currents that rotated in the same sense as the wind vector, as has previously been found in the literature. However, despite the comparable wind speeds while the storm was in western Pacific and SCS, the SST cooling was much more intense in SCS. The reason was because in SCS, the surface layer was thinner, the vorticity field of the Kuroshio was cyclonic, and moreover a combination of larger Coriolis frequency as the storm moved northward and the typhoon’s slower translational speed produced a stronger resonance between wind and current, resulting in strong shears and entrainment of cool subsurface waters in the upper ocean.

Appendix: Vertical wind mixing

We wish to compare the amount of energy required to mix a deep cool layer with the thin surface layer of South China Sea with that required to mix with the thicker surface layer of the western Pacific. That will show that when the initial mixed layer is thin, it is easier to lower its temperature through mixing with the cooler water in the deep layer. The following developments also re-derive Eq. 2 of Oey et al. (2006) in a different way.

The physical situation is illustrated in Fig. 12. Initially, warm fluid in the surface layer of thickness h ₁ and density ρ − δρ overlies cooler and denser fluid in the lower layer of thickness h ₂ and density ρ (left panel). Then wind mixes the two fluids into one layer of thickness h ₁ + h ₂ and density ρ _am (right). In this idealization, we assume that there is little agitation of the fluid below the lower layer. By mass conservation, the ρ _am is given by:

$$ {\rho}_{\mathrm{am}}=\left[\rho \times {h}_2+\left(\rho -\delta \rho \right)\times {h}_1\right]/\left[{h}_1+{h}_2\right] $$

(3)

Fig. 12

Left, two layers of different densities ρ and ρ − δρ (δρ > 0) before the onset of mixing (by wind). Right, wind mixes the fluids and homogenizes the two layers to one layer of some intermediate density = ρ _am

Wind pumps energy into the fluid by doing work through turbulent mixing, and thereby increases the potential energy of the fluid. The rate (per unit area) at which the wind does work is simply equal to the frictional stress (the drag force per unit area acted by the ocean surface on the air above) = ρ _a C _d |W|² times the air parcel’s speed (assume that the wind speed |W| >> ocean current speed), thus:

$$ \left(\mathrm{Wind}\;\mathrm{Work}\right)/\mathrm{Area}={\rho}_{\mathrm{a}}{C}_{\mathrm{d}}{\left|W\right|}^3\mathrm{in}\ \mathrm{J}/\left({\mathrm{m}}^2\ \mathrm{s}\right) $$

(4)

where ρ _a is the air density and C _d is the drag coefficient. If the wind blows for a time period = τ, then the energy (per unit area) supplied by the wind to the water column, or the power dissipation by the wind, is:

$$ \mathrm{P}\mathrm{O}\mathrm{W}=\gamma {{\displaystyle {\int}_0^{\tau, }{\rho}_{\mathrm{a}}{C}_{\mathrm{d}}\left|W\right|}}^3\mathrm{d}t,\kern0.37em \mathrm{in}\;\mathrm{J}/{\mathrm{m}}^2. $$

(5)

Here, the wind efficiency factor γ (<1) accounts for the fact that not all the wind work is used to produce mixing in the water column.

Wind work raises the potential energy (PE; per unit area) of the fluid. To see, we compare the PE after mixing, PE_AM, with the PE before mixing, PE_BM. The PE is simply the work done in raising the fluid through the water column:

$$ \begin{array}{c}\hfill {\mathrm{PE}}_{\mathrm{BM}}={\displaystyle {\int}_{-{h}_1-{h}_2}^0\rho^{\prime }gz\mathrm{d}z^{\prime }=}{\displaystyle {\int}_{-{h}_1-{h}_2}^{-{h}_1}\rho gz\mathrm{d}z^{\prime }}+{\displaystyle {\int}_{-{h}_1}^0\left(\rho -\delta \rho \right)gz\mathrm{d}z^{\prime }}\hfill \\ {}\hfill =-\left(g/2\right)\left[\rho {h}_2^2+\right.2\rho {h}_1{h}_2+\left(\rho -\delta \rho \right){h}_1^2\hfill \end{array} $$

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Similarly,

$$ {\mathrm{PE}}_{\mathrm{AM}}={\displaystyle {\int}_{-{h}_1-{h}_2}^0{\rho}_{\mathrm{am}}gz\mathrm{d}z^{\prime }=-\frac{g{\rho}_{\mathrm{am}}{\left({h}_1+{h}_2\right)}^2}{2}=-\frac{g\left({h}_1+{h}_2\right)\left[\rho {h}_2+\left(\rho -\delta \rho \right){h}_1\right]}{2}} $$

(7)

after using Eq. 3. The change in PE, δPE = PE_AM − PE_BM, is then given by:

$$ \delta \mathrm{P}\mathrm{E}=\left(g/2\right)\delta \rho \times {h}_2{h}_1,\kern0.37em \mathrm{in}\;\mathrm{J}/{\mathrm{m}}^2 $$

(8)

which is positive, confirming that the PE of the mixed fluid is indeed increased. The ratio,

$$ \varPhi ={\delta \mathrm{P}\mathrm{E}/\mathrm{P}\mathrm{O}\mathrm{W}=\left[g\delta \rho \times {h}_2{h}_1/2\right]/\gamma {\displaystyle {\int}_0^{\tau, }{\rho}_{\mathrm{a}}{C}_{\mathrm{d}}\left|W\right|}}^3\mathrm{d}t $$

(9)

is Eq. 2 of Oey et al. (2006). It is a ratio between the amount of energy required to thoroughly mix two fluid layers and the amount of wind energy available to do the mixing. In the case of typhoon Nuri, its intensity before and after entering South China Sea is approximately the same, so that the corresponding wind power dissipation is also approximately equal. Taking the 26 °C isotherm as the base of the surface layer, then h _1SCS ≈ 40 m in South China Sea, and h _1WP ≈ 70 m in the western Pacific (see Fig. 7a). Assuming that h ₂ >> h ₁, we then have from Eq. 9:

$$ {\varPhi}_{\mathrm{WP}}/{\varPhi}_{\mathrm{SCS}}\approx {h}_{1\mathrm{W}\mathrm{P}}/{h}_{1\mathrm{S}\mathrm{C}\mathrm{S}}\approx 2, $$

(10)

and it takes about half of the wind energy to mix the thinner surface warm layer in South China Sea than in western Pacific.