On Saturday, June 8, 2013 8:45:12 AM UTC-4, Andy Blackburn wrote:
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:



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


I started to work on the equation the other night, but the Stanley Cup playoffs got me distracted :-) It involved dusting off my old textbooks, but I think Andy's work passes the sniff test.

Couple of points: Starting temp is obviously important. 2C vs 15C makes a difference. Out West, I'm sure most flights start with temps on the ground well up into the 20C range, so it's probably not a factor. Back East, we do a lot of ridge flights where the surface temps are probably less than 5C. If the water sat in a car-top container overnight or even in the wings, it's likely that the water started off around there. If it came out of a spigot, it's likely that it was closer to 10C.

Anecdotally: There were a lot of long ridge flights from Blairstown in the 1990s. Temps at altitude probably -5C and ground temps just over 0. Typical flights were 5-6 hours (any more, and it was the pilot freezing solid that was the issue). I don't recall anyone having any issues with bags or tanks freezing into solid blocks. I do remember a couple of cases where leaky LS valves lead to freezing on the flaps or tail. I also believe there were a couple of cases of asymmetrical dumping due to one wing valve freezing. The tail one was kinda scary, because I believe the rudder basically jammed.

So, my take is that there are significant risks, especially if the lower levels where you are flying are at or below freezing. But, the risks aren't so much the solid block of ice as problems with the valves and freezing from any small leaks.

P3