wrote:
If you had a supertanker dispersing the coolant you could have it
follow along within the eye or it could precede ahead of it within its
predicted track.

Any other technical obstacles aside, the moment the supertanker in the
eye was unable to keep up with the storm's forward motion, it would be
S.O.L. (short on luck).

"Uncle Al" wrote in message
...
[snip crap]

Psychotic ineducable boring spammer Alan Schwartz,
the royal ****wit, "Uncle Al"
mumble some crap in message

...
Why are you having so much trouble with basic algebra?
Let L_1 = distance light travels in going from Sam to Joe, as
measured in the stationary frame.
1) L_1 = cL/(c-v)

What a right royal stooopid mother****er.
See the peeing puppy moortel, he'll not be glad to add
you to his list of truly IMMORTAL fumbles. I will, though.

[quote]
we establish by definition that the "time" required by a turtle to
travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

[quote]
For velocities greater than that of a turtle our deliberations become
meaningless; we shall, however, find in what follows, that the velocity
of a turtle in our theory plays the part, physically, of an infinitely
great velocity.
[quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

Nothing can go faster than a turtle.

Oops!... Did I say 'a turtle'? Sorry...'light'.

Androcles

"Uncle Al" wrote in message
...
[snip crap]

The Chinee told you to **** off, eh?
Does it burn, stooopid, does it burn?

Psychotic ineducable boring spammer Alan Schwartz,
the royal ****wit, "Uncle Al"
mumble some crap in message

...
Why are you having so much trouble with basic algebra?
Let L_1 = distance light travels in going from Sam to Joe, as
measured in the stationary frame.
1) L_1 = cL/(c-v)

What a right royal stooopid mother****er.
See the peeing puppy moortel, he'll not be glad to add
you to his list of truly IMMORTAL fumbles. I will, though.

[quote]
we establish by definition that the "time" required by a turtle to
travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

[quote]
For velocities greater than that of a turtle our deliberations become
meaningless; we shall, however, find in what follows, that the velocity
of a turtle in our theory plays the part, physically, of an infinitely
great velocity.
[quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

Nothing can go faster than a turtle.

Oops!... Did I say 'a turtle'? Sorry...'light'.

Androcles

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!

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!

Borek wrote:
On Sun, 25 Sep 2005 01:51:45 +0200, wrote:

26º C surrounding temperatures. I'll take as a guess for the thickness
of 1 cm. Then this is a volume of 10,000m x 10,000m x .01m = 1,000,000
m^3. This is 1,000,000 metric tons of water. Then it would require that
amount in weight of NH4NO3. The worldwide production of ammonium


Adding such an amount of fertilizer to sea water is asking for troubles.
It is just like fertilizers washed from fields being dangerous for lakes.

Best,
Borek

Yup, with that amount of fertilizer, the resulting algae bloom will kill
all the fish within hundredss of miles.

wrote:
...
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.
.
.
.
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.



You see that the heat flux has the thickness in the denominator while
the amount of heat is linearly dependent on mass which is density *
area * thickness. So the amount of heat has thickness in the numerator.
Then when you calculate the amount of time by dividing the amount of
heat by the heat flux you see this is dependent on the square of the
thickness. So a layer only 10 cm, 4 inches, thick would last 10^2 = 100
times as long as a 1 cm thick layer or 23.5*100 = 2350 minutes, about
40 hours.
Another surprising thing is that the temperature difference DeltaT
cancels out when you calculate the length of time. So it wouldn't get
better by making the layer cooler.



- Bob

A Espinoza wrote:
Borek wrote:
On Sun, 25 Sep 2005 01:51:45 +0200, wrote:

26º C surrounding temperatures. I'll take as a guess for the thickness
of 1 cm. Then this is a volume of 10,000m x 10,000m x .01m = 1,000,000
m^3. This is 1,000,000 metric tons of water. Then it would require that
amount in weight of NH4NO3. The worldwide production of ammonium


Adding such an amount of fertilizer to sea water is asking for troubles.
It is just like fertilizers washed from fields being dangerous for lakes.

Best,
Borek

Yup, with that amount of fertilizer, the resulting algae bloom will kill
all the fish within hundredss of miles.

If you are going to use a cooling reaction (there are several besides
that of ammonium nitrate) you may want to keep the material insides
packets as done with instant cooling packs. For example on the page I
cited, this was the method suggested to produce ice when the
electricity was out:

Making ice without machinery
http://www.madsci.org/posts/archives...5573.Ch.r.html

In this case of course you don't want the drinkable water to come in
contact with the coolant material.
You would need to connect these packets together with strong light
fibers so after use they could be collected so as not to create a
pollution problem themselves.


Bob Clark

Found a web page for calculating the degree of cooling that can be
achieved by using ice in warm water:

Cooling a Cup of Coffee.
http://hyperphysics.phy-astr.gsu.edu...oocof2.html#c3

It uses the equation for calculating the amount of heat gained or lost
versus the temperature change according to the specific heat of the
material: heat gained or lost = (specific heat)*mass*(delta
temperature). It notes that when using ice, you also have to take into
account the extra heat lost for the phase change from ice to liquid,
the latent heat of fusion, 80 cal/gm for water:

"Cooling a Cup of Coffee
You have a 200 gram cup of coffee at 100ºC, too hot to drink. How much
will you cool it by adding 50 gm of ice at 0ºC?

Heat lost by coffee = Heat gained by ice

-Qcoffee = Qice

-c*mc*delta-Tcoffee = mi*Lf +c*mi*deltaTice

(1 cal/gm ºC)(200 gm)(100-Tf) = (50 gm)(80 cal/gm) + (1 cal/gm ºC)(50
gm)(Tf -0)

20,000 - 200*Tf = 4,000 +50*Tf

(20,000-4,000)/250 = Tf =64ºC"

The calculation is saying the coffee and the water(ice) are going to
be brought to some common temperature Tf. The heat gained by the ice is
first the latent heat of fusion to melt the ice, then on top of that
there is heat gained to bring the now liquid water from 0ºC to Tf.

I want to adapt this to a case where the ice is introduced at a much
lower temperature. I'll use -200ºC. This is a slightly lower than the
temperature for liquid nitrogen and is easily achieved with
refrigeration techniques. Let mi be the mass of the ice, and let the
mass of water you want to cool be larger by a factor of k, so k*mi.
I'll let the water be originally at 25ºC. I'll need as well the
specific heat of ice. This is actually about half the specific heat of
liquid water, .5 cal/gm ºC. Then on the ice side of the equation, I'll
calculate the heat gained by the ice in bringing it from -200ºC to
0ºC, plus the latent heat of fusion in melting the ice, plus the heat
gained in bringing the former ice now liquid water from 0ºC to the
final temperature Tf. And on the water side, I'll just have the heat
lost in bringing the water down from 25ºC to the final temperature Tf:

(1 cal/gm ºC)*k*mi*(25-Tf) = (.5 cal/gm ºC)*mi*200 + mi*(80 cal/gm) +
(1 cal/gm ºC)(Tf-0)

The mi cancels out to give:

k*(25-Tf) = 100 + 80 + Tf = Tf + 180.

So k = (Tf + 180)/(25 - Tf). This gives how much more water can be be
chilled to the temperature Tf starting with the water at 25ºC and the
ice at -200ºC. Then for a 5ºC temperature drop for the water this
would be Tf = 20ºC and k = 200/5 = 40, i.e., the ice could chill 40
times as much water as the ice carried. For a 100,000 ton capacity
container ship this would correspond to 4,000,000 tons of water chilled
or a 20km by 20 km by 1 cm thick volume or a 10km by 10km by 4cm thick
volume.
However, a problem would be insuring that the ice was only chilling
the surface layer for the greatest surface area to be chilled. Perhaps
the ice could be distributed in thin flakes or small cubes so that it
would melt quickly near the surface.



Bob Clark


wrote:
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!