Thanks for the response. It occurs to me that for the ice scenario if
I'm going to use the same amount of ammonium nitrate for the same
amount of water to be frozen, I might as well just transport the ice.
Would this hold as well for just lowering the temperature rather than
freezing? That is if 30,000 tons of ammonium nitrate would lower
1,000,000 tons of water 5 degrees F, then would 30,000 tons of ice at
freezing also lower the temperature of 1,000,000 tons of water by 5
degrees F? Can you calculate this case?
If so, or if the amounts are comparable, then this might provide a
more environmentally benign means of accomplishing the same thing.
There are cargo transport ships called container ships that can
transport up to 100,000 tons of cargo in standard-sized containers.
They can travel in the range of 50 km/hr. Here's an image of one:

LIONS GATE BRIDGE.
http://www.shipphoto.net/lions%20gate%20bridge.htm

These containers are often refrigerated for transporting perishables.
Then we would freeze the ice ahead of time (it would take time to
freeze this much ice) and keep the ship on dock fully loaded with ice
until needed.
At 50 km/hr the ship could travel 1200 km in 24 hours. We usually have
2 to 3 days warning of when a hurricane will hit. So these ships should
be able to intercept the hurricane before they reach land.
As you see from the photo, these ships are rather unwieldly, so it's
unlikely they could survive travelling through the high winds to dump
the ice within the hurricane eye. We would have to use the method of
placing the ice in front of the planned track of the hurricane.
We can probably increase the effectiveness by lowering the temperature
of the ice even further. For example, the temperature of liquid
nitrogen at 77 K is easily achieved and maintained with refrigeration.
Then we can keep the water ice in the containers at this temperature.
This is a factor of 3.5 lower than the freezing point of water on the
Kelvin scale so the amount of ocean water whose temperature we can
lower should also be increased by that factor.
The advantage of this proposal is that the container ships with
refrigerated containers are already in operation so we could implement
this like tomorrow. One problem though is that the heavy lift cranes
for moving these containers are kept on shore, not on ship. So we would
need to have an accompanying ship or ships with heavy crane capability.


Bob Clark



wrote:
Alright Mr. Clark - let's think about heat transfer for a bit here...
let's pretend there is no mixing in the ocean and that everything is
perfectly calm during a hurricane (excellent assumptions no?). Alright
now we have a 1 cm think layer of water at 20C and the rest of the
depth at 25C (compared to the 1 cm that is 20C, the amount of water
below it at 25C might as well be infinite for this little exercise).
I will assume no heat is transferred from air to water (another
excellent and good assumption!). Let's see how fast we lose this
critical temperature difference.

The temperature at the surface of the ocean is the most critical so we
can think of this cool layer as trying to maintain 20C at the surface
of the ocean. Therefore, the DeltaT is 5C, the difference in T between
the bulk ocean and the surface.

The thermal conductivity of water is roughly 0.6J/(s-m-C). The
thickness of the layer is Delta-x = 0.01m. We will consider this
problem using unit area of 1m^2 (we will assume this is an infinite
plane of cool water, a reasonable assumption for an area somewhere not
at the edge of your cooled ocean area).

The heat flux is given by Q = (k/Delta-x)*A*DeltaT = (0.6/0.01)*1*5 =
300 J/s. (you can check it yourself, the units work out propoerly - i
was lazy and didnt want to type them again).

Alright so we have heat flowing at 300 J/s. As the temperature
difference drops (i.e. the temp of the surface rises), Q also drops
linearly with the drop in DeltaT. Let us use 150 J/s as an average
over a large part of this temperature rise at the surface.

The amount of heat it takes to raise the temp of that water by 5C is
given by (specific heat)*(mass)*(deltaT)

energy needed = 4.19 J/(g-C)*(.01m*1m^2*1000000g/m^3)*(5C) = 209500 J

Now at 150 J/s this says our water will be roughly heated in 209500/150
s =~ 1400 s =~ 23.5 min.

So you will loose your deltaT in less than 25 min assuming perfectly
calm water, no heat transfer through the air, no mixing in the water,
and my crazy simple model that assumes your 1 cm heats identically all
the way through (as the water in that 1 cm layer heats, the surface
will heat more and more rapidly as the water near it warms - and finite
element model [or finite difference] here would help alot). Lets think
about a hurricane for a moment he huge waves, extremely turbulent
water, high winds whipped the water around - the mixing in the water
itself would kill your cool layer in a matter of seconds (when you put
ice cubes in a container with water and shake it around, you will melt
the ice faster than if you take that same volume of water, put the same
amount of ice in it and heat it over a flame). I'd say plan on cooling
down to a rather deeper depth than 1 cm - like maybe a couple hundred
feet and pray that they are no currents down there to sweep your cool
water a couple hundred miles away.

Let me know how it goes - and don't kill ALL the fish if you help it.
And give me a heads up - if we are going to do this, let me set up my
ammonium nitrate plant!