Jim wrote:
On Friday, June 7, 2013 8:53:49 AM UTC-4, wrote:
On Wednesday, June 5, 2013 8:41:20 AM UTC-7, Matt Herron Jr. wrote:

Can anyone share some wisdom on using water at high elevations for long
durations? How do you know your fin or wing tanks will not freeze? If
I am at 18K for 6 hrs in the Sierras, I really don't want my vertical
stab splitting open in flight. Has this ever happened? Any guidance
would be appreciated.







Matt





What you're basically looking for is a solution to a reasonably
straight-forward heat transfer problem. We can make a set of assumptions
to make this a solvable problem.



1) The temperature of the water is basically spatially uniform.

2) The temperature of the surrounding atmosphere doesn't change.

3) The fluid properties (density and specific heat) are constant.

4) The heat transfer is convective only (from cold air flowing around
the water container). No conduction or radiation.

5) There's no forced convection (i.e. no fans or forced flow into the ballast tank)

6) The free convection heat transfer is fairly efficient or the cold air
motion outside of the water container is substantial (i.e. Gr and Pr are large)



If you do this, you end up with an equation that looks like this:



-h * A * (T - To) = rho * V * (Cp * dT + hfs) / dt



h = the convective heat transfer coefficient, which basically measures
how efficiently heat is being transferred. With our assumptions, we can
roughly approximate h = 1.52 * (T - To)^(1/3).

A = the surface area of the water container

T = the temperature of the water at time t, We'll use 32 °F here since
most of the heat transfer will be occuring while the water is undergoing the phase change

To = the outside temperature (this needs to be below 32 °F)

rho = the density of the water

V = the volume of the water

Cp = the specific heat of the water (assume around 4.2 kJ/ kg K)

dT = the difference between the initial and final water temperatures

dt = the time it takes to freeze

hfs = the enthalpy of fusion (about 334 kJ/kg)



Rearrange the equation and we set the energy required to freeze the
water (hA(T-To)) is equal to the energy required to cool it
(rhoVCpDT/dt) plus the energy required to turn it into a solid (rhoVhfs/dt).

Suppose you have a typical water bottle as a simple example.



Starting with h, if you begin at room temperature, h = 1.52 * (25-0)^(1/3) = 4.4 (W/m^2 K)



Assume a cylindrical container meauring about 6" x 2.5". So PI * D * L = 0.03 (m^2)



The outside air is at 0 °F (-18 °C) and the water starts at about room
temperature (20 °C). So T - To = -40 (°C)



The volume is about 500 mL, the density is about 1000 kg/m3, so the mass is about 0.5 kg.



The specific heat we said above was 4.2 (kJ/kg K) or 4200 (J/kg)



The temperature change will be from room temperature to freezing, so 20 °C.



The enthalpy of fusion from above is 334,000 J/kg.



Plugging all this in and solving for dt gives a time of 31,667 s or
about 8 hr. 48 mins. For any volume of room temperature water going into
a freezing environment, the only thing that will change will be the
surface area (A), and the volume (V). The rest you can keep the same as
a basic approximation so you have:



t = 1,900,000 V/A



where V and A are in m^3 and m^2, respectively.



The V/A for tail tanks and wing tanks vary only slightly. We should
probably adjust the heat transfer coefficient downward a bit from the
above example since the tanks in most gliders are insulated with the
composite/foam sandwich of the structure. This will make the time
longer. If it's warmer than 0 °F it will take longer still, so for
normal thermal soaring you would expect never to get to frozen solid.
Maybe on a really long wave flight you should worry.



Lastly, as has been pointed out, depending on your venting, valving and
CG considerations, you may have localized water management issues from
small-scale freezing, but I wouldn't worry about a giant block of ice
exploding from my wing or tail.



9B

Interesting calculation. I wonder why use such assumptions as
"temperature of the surrounding atmosphere doesn't change", which it does
as altitude changes and weather changes through a long flight. Also,
"room temperature" is hardly close when filling tanks in winter or from
well water..How does the result look if these assumptions are at their worst maximum?

-Jim


--
9B

Jim

The assumption is the temperature of the atmosphere doesn't change due to
the heat transfer from your ballast water. I think that's a pretty safe
assumption given the relative heat capacities of the earth's atmosphere
versus 40 gallons of water. Otherwise ballasted sailplanes would be
responsible for global warming. ;-)

You are correct that over the course of a flight you go up and down in
altitude. You are also correct that the starting temperature for ballast
water is probably not room temperature. I picked 0 degrees for the OAT even
though the more typical temperature for 18,000' is closer to 20 degrees and
is considerably warmer at lower altitudes. I would say the temperature
differential assumption is actually pretty aggressive. I barely see
freezing temps on most flights well up into supplemental oxygen altitudes.
You also have to appreciate that the phase change to ice is the thing that
consumes the most energy, not getting from room temp to 55 degrees. The
simple sensitivity analysis I did would indicate that the actual time to
freeze your ballast tanks solid is much longer than 9 hours under any
realistic scenario.

About the time I made my original post I also put a gallon of water in my
zero degree freezer - I'll let you know how that goes...

9B