Here’s a beautiful photo of Hurricane Irene as it makes its way up the East Cost this weekend (tomorrow is Saturday, Sunday comes afterwards).

There have been a few hair-brained schemes to mitigate the intensity of a hurricane throughout the years. I say “hair-brained” because it seems highly unlikely that either of these could potentially affect the path or strength of a hurricane. It’s simply a question of the amount of energy produced by a hurricane (which is a lot) compared with the energy released by giant blocks of ice and nuclear warheads (which are comparatively small). Those are the hurricane mitigation attempts that I’d like to focus on. Because while they may be “hair-brained” they can prove to be instructive about environmental science, geometry, and calculus.

**Artifacts**

- Mitigating hurricane damage by
- Using giant blocks of ice to cool the sea surface temperature (SST).
- Detonating a nuclear bomb inside a hurricane.

**Guiding questions**

- How big would a block of ice have to be to cool the SST in front of a hurricane by, let’s say, one degree Celsius?
- Wouldn’t a lot of that ice melt on its way down from the arctic?
- How much energy does an atomic bomb have compared to a hurricane?

**Suggested activities**

- Research the amount of energy in a hurricane.
- Calculate the distance from the arctic to the Atlantic hurricane basin, and speed at which a naval vessel could transport the gigantic block of ice.
- Research the amount of energy released by a nuclear bomb.
- Have groups decide which method is a more feasible achievement.

**Potential solutions**

Well, let’s look at the energy of a hurricane. Take it away, NOAA.

One can look at the energetics of a hurricane in two ways:

- the total amount of energy released by the condensation of water droplets or …
- the amount of kinetic energy generated to maintain the strong swirling winds of the hurricane (Emanuel 1999).
It turns out that the vast majority of the heat released in the condensation process is used to cause rising motions in the thunderstorms and only a small portion drives the storm’s horizontal winds.

Method 1) – Total energy released through cloud/rain formation:An average hurricane produces 1.5 cm/day (0.6 inches/day) of rain inside a circle of radius 665 km (360 n.mi) (Gray 1981). (More rain falls in the inner portion of hurricane around the eyewall, less in the outer rainbands.) Converting this to a volume of rain gives 2.1 x 10^{16}cm3/day. A cubic cm of rain weighs 1 gm. Using the latent heat of condensation, this amount of rain produced gives5.2 x 10^{19}Joules/day or

6.0 x 10^{14}Watts.

This is equivalent to 200 times the world-wide electrical generating capacity – an incredible amount of energy produced!Method 2) – Total kinetic energy (wind energy) generated:For a mature hurricane, the amount of kinetic energy generated is equal to that being dissipated due to friction. The dissipation rate per unit area is air density times the drag coefficient times the windspeed cubed (See Emanuel 1999 for details). One could either integrate a typical wind profile over a range of radii from the hurricane’s center to the outer radius encompassing the storm, or assume an average windspeed for the inner core of the hurricane. Doing the latter and using 40 m/s (90 mph) winds on a scale of radius 60 km (40 n.mi.), one gets a wind dissipation rate (wind generation rate) of1.3 x 10^{17}Joules/day or

1.5 x 10^{12}Watts.

This is equivalent to about half the world-wide electrical generating capacity – also an amazing amount of energy being produced!

That’s a lot of joules/day, easily dwarfing the release of a (and several!) nuclear warheads. As for the giant block of ice…

Well, the **specific heat of water** is about 4 joules/gram, which means that it would take 4 joules of energy to raise or lower one gram of water by one degree Celsius. Water weighs about 1000 kg per cubic meter. If you want to cool a swath of ocean, say, 3000 sq km to a depth of two meters, then that’s,

3000 x 2 x 1000 = 6 million kg of water

It would take a lot of ice to cool that much water by a single degree. And hurricanes span a lot more than 3000 sq. km.

But for this instance, you would need 24,000,000,000 joules of energy from the ice.

The latent heat of ice is 334 kJ/kg. So my back-of-the-envelope calculation (this is all very scientific) tells me that we need

24,000,000,000 J x 1 kJ/1000 kJ x 1 kg/334 kJ = 12,000,000 kg of ice

And that’s at the point of the hurricane. And that’s one degree. How big is 12,000,000 kg of ice, anyway? Another, possibly more interesting more interesting question is a calculus problem: how much ice would melt before we got it to the hurricane. Better yet: *how much ice would we have to start with in order to have 12,000,000 kg of ice by the time we get to the Atlantic?*

Oh my ice-melting/calculus sense is tingling!

There are 3 trillion sq meters in 3,000 sq/km. Multiply 3 trillion by 2 m (depth) times 1000 kg. Therefore isn’t the mass of that water 6 quadrillion kg of water?