freezing water in space
byblow
or bladder) takes a long time to freeze in space because convection
and conduction of heat can't occur in a vacuum. But surely water in a
tank or bladder will eventually freeze in space if shaded from the
sun, right?
Can anyone tell me how long it would take? The water is distilled, so
no worries about impurities. I'm interested in how long it would take
for large amounts to freeze, like 1,000 tonnes or more. Does the
amount of water affect how long it takes to freeze?
Jeff Findley
In article <80d64180-bb17-4eb8-a5d5-f1428de554b0
@t9g2000vbs.googlegroups.com>, pstu...@gmail.com says...
>
> sun, right?
>
This is a chemistry and thermodynamics question. If you dig a bit, I'm
sure you can find the relevant equations.
Note that the time it takes will depend on a lot of things, like how
much water is in the container (i.e. its mass), the initial starting
temperature of the water, what the container is made of, how thick the
container is, the surface area of the container, and etc.
Jeff
--
" Solids are a branch of fireworks, not rocketry. :-) :-) ", Henry
Spencer 1/28/2011
Steve Willner
byblow <pstu...@gmail.com> writes:
> A friend of mine who's a physicist says water in a container (a tank
> or bladder) takes a long time to freeze in space because convection
> and conduction of heat can't occur in a vacuum.
There are two issues: conducting heat from the water to the container
wall, and radiating heat from the container wall into space. The
latter is easy enough to calculate: a perfect emitting surface at 0
deg_C radiates just over 300 W m^-2. Real materials aren't perfect
emitters, but you can get to 0.8 or better pretty easily. Freezing
water into ice takes 334 J/kg (according to Wikipedia, which is the
first source I found), so it's easy to work out a minimum time for
any amount of water, given the tank shape.
If there's poor conduction from the water in the interior to the
container wall, the actual time will be longer. You could look up
the thermal conductivity of water and work it out. If rapid freezing
is needed, you could use a tank that is small in at least one
dimension so no part of the water is far from a tank wall or put
copper filaments or something inside the tank to improve conduction.
If you try this for real :-), make sure your container can
accommodate the expansion as the water freezes.
> Does the amount of water affect how long it takes to freeze?
Yes: heat loss goes as the surface area, but the enthalpy of fusion
goes as the mass (or volume) of water. As noted above, tank shape is
important, too. A small tank can have pretty much any shape, but it
won't be easy to make a large tank with an odd shape nor keep it
shaded from the Sun if you can make it.
--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
Cambridge, MA 02138 USA
Keith Henson
The question is how long does it take radiation to take away the neat
of water freezing?
Starting with 1000 tons, water is a ton per cubic meter so this would
be 1000 cubic meters, or ten by ten by ten meters as a cube.
The cube surface area would be be 600 square meters. At the freezing
point of water, the black body radiation is ~300 W/m^2 or J/s/m^2, so
the radiation from the whole thing would be around 180,000 J/s
It takes 3.34 x 10^5 J/kg to freeze water. At 1 M kg that's 3.34 x
10^11 J. Divided by the radiation rate of 1.8 x 10^5J/s, 1.86 x 10^6
seconds, 515 hours or 21.5 days.
A cubic meter of water would freeze faster. It would lose heat at 6 x
300 or 1800 j/s and, to freeze need to lose (3.34 10^5 j/kg x 1000 kg)/
(1.8 x 10^3)/s or 1.86 x 10^5 s or 2.1 days.
This doesn't take into account water cooling to freezing or the
conduction of heat out through the ice, but I expect them to be minor
corrections especially for the 1000 ton case.
Keith
byblow
Wow! I had no idea it was so complicated. Could I impose upon you to
calculate the time it would take for 50,000 tonnes of water in a
spherical bladder -- like a water balloon -- to freeze in space
(assuming it's shaded)?
Keith Henson
Sheesh. I am sure you took geometry long after I did.
V (4/3)*pi*r^3
50,000 /(4/3)*pi r^3
r 22.85 m
A 4 * pi * r^2
A 6563 m^2
heat loss A * 300 j-s/m^2
q 1.969 M J-s for the whole thing.
>From above, you need 3.34 * 10^11 * 50 / q
T 8481392 s
2356 hr
98 days
Again, no correction for the conductivity of ice and the water is at
freezing to start. And I have made no effort to check the results.
Keith