The Rise of the Ocean Due to a Hurricane A hurricane is a tropical storm formed over the ocean by low atmospheric pressures. As a hurricane approaches land, inordinate ocean swells (very high tides) accompany the hurricane. A Class-5 hurricane features winds in excess of 250 km/h, although the wind

Question

The Rise of the Ocean Due to a Hurricane A hurricane

is a tropical storm formed over the ocean by low atmospheric pressures. As a hurricane approaches land, inordinate ocean swells (very high tides) accompany the hurricane. A Class-5 hurricane features winds in excess of 250 km/h, although the wind velocity at the center “eye” is very low.
Figure 5–43 depicts a hurricane hovering over the ocean swell below. The atmospheric pressure 320 km from the eye is 762 mm Hg (at point 1, generally normal for the ocean) and the winds are calm. The atmospheric pressure at the eye of the storm is 560 mm Hg. Estimate the ocean swell at (a) the eye of the hurricane at point 3 and (b) point 2, where the wind velocity is 250 km/h. Take the density of seawater and mercury to be 1025 kg/m³ and 13,600 kg/m³, respectively, and the density of air at normal sea-level temperature and pressure to be 1.2 kg/m³.

Accepted Answer

A hurricane is moving over the ocean. The amount of ocean swell at the eye and at active regions of the hurricane are to be determined.
Assumptions 1 The airflow within the hurricane is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). (This is certainly a very questionable assumption for a highly turbulent flow, but it is justified in the discussion.) 2 The effect of water sucked into the air is negligible.
Properties The densities of air at normal conditions, seawater, and mercury are given to be 1.2 kg/m³, 1025 kg/m³, and 13,600 kg/m³, respectively.
Analysis (a) Reduced atmospheric pressure over the water causes the water to rise.
Thus, decreased pressure at point 2 relative to point 1 causes the ocean water to rise at point 2. The same is true at point 3, where the storm air velocity is negligible. The pressure difference given in terms of the mercury column height is expressed in terms of the seawater column height by

[latex]\Delta P = (\rho gh)_{Hg} = (\rho gh)_{SW} → h_{SW} = \frac{\rho _{Hg}}{\rho _{SW}} h_{Hg}[/latex]

Then the pressure difference between points 1 and 3 in terms of the seawater column height becomes

[latex] h_3 = \frac{\rho _{Hg}}{\rho _{SW}}h_{Hg} = (\frac{13,600 kg/m^3}{1025 kg/m^3} ) \left[(762 - 560) mm Hg\right] \left(\frac{1 m}{1000 mm} \right) = 2.68 m[/latex]

which is equivalent to the storm surge at the eye of the hurricane (Fig. 5–44) since the wind velocity there is negligible and there are no dynamic effects.

(b) To determine the additional rise of ocean water at point 2 due to the high winds at that point, we write the Bernoulli equation between points A and B, which are on top of points 2 and 3, respectively. Noting that [latex]V_B ≅ 0[/latex] (the eye region of the hurricane is relatively calm) and [latex]z_A = z_B[/latex] (both points are on the same horizontal line), the Bernoulli equation simplifies to

[latex] \frac{P_A}{2 g} + \frac{V^2_A}{\rho g} + \cancel{z_A} = \frac{P_B}{\rho g} + \overset{0}{\frac{\cancel{V^2_B}}{2g} } + \cancel{z_B} → \frac{P_B - P_A}{\rho g} = \frac{V_A^2}{2g} [/latex]

Substituting,

[latex] \frac{P_B - P_A}{\rho g} = \frac{V_A^2}{2g} = \frac{(250 kg/h)^2}{2(9.81 m/s^2)} \left(\frac{1 m/s}{3.6 km/h} \right)^2 = 246 m[/latex]

where 𝜌 is the density of air in the hurricane. Noting that the density of an ideal gas at constant temperature is proportional to absolute pressure and the density of air at the normal atmospheric pressure of 101 kPa ≅ 762 mm Hg is 1.2 kg/m³, the density of air in the hurricane is

[latex] \rho _{air} = \frac{P_{air}}{P_{atm air}} \rho _{atm air} = \left(\frac{560 mm}{762 mm} \right) (1.2 kg/m^3) = 0.882 kg/m^3[/latex]

Using the relation developed above in part (a), the seawater column height equivalent to 246 m of air column height is determined to be

[latex] h_{dynamic} = \frac{\rho _{air}}{\rho _{SW}} h_{air} = \left(\frac{0.882 kg/m^3}{1025 kg/m^3} \right)(246 m) = 0.21 m[/latex]

Therefore, the pressure at point 2 is 0.21 m seawater column lower than the pressure at point 3 due to the high wind velocities, causing the ocean to rise an additional 0.21 m. Then the total storm surge at point 2 becomes

[latex]h_2 = h_3 + h_{dynamic} = 2.68 + 0.21 = 2.89 m[/latex]

Discussion This problem involves highly turbulent flow and the intense breakdown of the streamlines, and thus the applicability of the Bernoulli equation in part (b) is questionable. Furthermore, the flow in the eye of the storm is not irrotational, and the Bernoulli equation constant changes across streamlines (see Chap. 10). The Bernoulli analysis can be thought of as the limiting, ideal case, and shows that the rise of seawater due to high-velocity winds cannot be more than 0.21 m.
The wind power of hurricanes is not the only cause of damage to coastal areas. Ocean flooding and erosion from excessive tides is just as serious, as are high waves generated by the storm turbulence and energy.

Question 5.9: The Rise of the Ocean Due to a Hurricane A hurricane is a tr...

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