fin/wing tanks freezing

The fix was to put a pint of RV antifreeze in the tail tank. RV antifreeze is propylene glycol and is safe even if you drink a little.

DLB


Dale, My significant other suggested I use White Zin. I countered with Gewurztraminer. Since it is not fit for human consumption, there is no conflict in pulling the dump actuator ;-)

Craig R.

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

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

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

"Craig R." wrote:
The fix was to put a pint of RV antifreeze in the tail tank. RV
antifreeze is propylene glycol and is safe even if you drink a little.

DLB


Dale, My significant other suggested I use White Zin. I countered with
Gewurztraminer. Since it is not fit for human consumption, there is no
conflict in pulling the dump actuator ;-)

Craig R.

Just watch which vintage of wine you're using. In the late 80s there was a
major scandal with various producers of Spättlese spiking the wine with
ethylene glycol. ;-)

Pete

Not having studied thermal dynamics, etc, I have two rookie questions. How does the container material (thin plastic "milk jug" in freezer vs layered fiberglass for tail tank) and the shape of the container (minimal surface area of a milk jug vs long thin tail) effect these calculations? ... also, head space in jug is zip and in tail tank is large....

On Saturday, June 8, 2013 8:04:15 PM UTC-7, Craig R. wrote:
Not having studied thermal dynamics, etc, I have two rookie questions. How does the container material (thin plastic "milk jug" in freezer vs layered fiberglass for tail tank) and the shape of the container (minimal surface area of a milk jug vs long thin tail) effect these calculations? ... also, head space in jug is zip and in tail tank is large....

Having not studied thermodynamics you have saved yourself from a lifetime of boredom (even it if was only an hour a week). To a good approximation, the shape is accounted for in the surface to volume estimate, and the head space is not consequential to heat exchange.

On Saturday, June 8, 2013 8:23:00 PM UTC-7, jfitch wrote:
On Saturday, June 8, 2013 8:04:15 PM UTC-7, Craig R. wrote:

Not having studied thermal dynamics, etc, I have two rookie questions. How does the container material (thin plastic "milk jug" in freezer vs layered fiberglass for tail tank) and the shape of the container (minimal surface area of a milk jug vs long thin tail) effect these calculations? ... also, head space in jug is zip and in tail tank is large....



Having not studied thermodynamics you have saved yourself from a lifetime of boredom (even it if was only an hour a week). To a good approximation, the shape is accounted for in the surface to volume estimate, and the head space is not consequential to heat exchange.

I looked up the effect of various materials on heat transfer and it would appear that 1/4" of foam-composite sandwich roughly halves the heat transmission, extending the time to freeze by roughly double.

The starting temperature of the water doesn't matter that much in the end result (if the end result you are looking for is a solid block of ice). The enthalpy of fusion (turning 32-degree water into 32-degree ice) represents 75-90 percent of the heat transfer versus 10-25 percent to take the water from its starting temperature to 32 degrees.

Again, you would have to be on one of those 16-hour, 25,000 foot Sierra wave flights to worry about frozen-solid ballast tanks - even then I remain a bit skeptical that you'd have a problem. The main risk on a typical summer thermal flight in the 16-18,000' range is from leaking water freezing up on a valve or hinge that you care about. People who have experienced this seem to report pretty consistently that you can break free most of the time. I probably would avoid dumping ballast in freezing conditions if possible.

9B

On Saturday, June 8, 2013 8:50:02 PM UTC-7, wrote:
On Saturday, June 8, 2013 8:23:00 PM UTC-7, jfitch wrote:

On Saturday, June 8, 2013 8:04:15 PM UTC-7, Craig R. wrote:



Not having studied thermal dynamics, etc, I have two rookie questions.. How does the container material (thin plastic "milk jug" in freezer vs layered fiberglass for tail tank) and the shape of the container (minimal surface area of a milk jug vs long thin tail) effect these calculations? ... also, head space in jug is zip and in tail tank is large....







Having not studied thermodynamics you have saved yourself from a lifetime of boredom (even it if was only an hour a week). To a good approximation, the shape is accounted for in the surface to volume estimate, and the head space is not consequential to heat exchange.



I looked up the effect of various materials on heat transfer and it would appear that 1/4" of foam-composite sandwich roughly halves the heat transmission, extending the time to freeze by roughly double.



The starting temperature of the water doesn't matter that much in the end result (if the end result you are looking for is a solid block of ice). The enthalpy of fusion (turning 32-degree water into 32-degree ice) represents 75-90 percent of the heat transfer versus 10-25 percent to take the water from its starting temperature to 32 degrees.



Again, you would have to be on one of those 16-hour, 25,000 foot Sierra wave flights to worry about frozen-solid ballast tanks - even then I remain a bit skeptical that you'd have a problem. The main risk on a typical summer thermal flight in the 16-18,000' range is from leaking water freezing up on a valve or hinge that you care about. People who have experienced this seem to report pretty consistently that you can break free most of the time. I probably would avoid dumping ballast in freezing conditions if possible.



9B

Okay - I put a 2 liter bottle of fresh tap water in my 0-degree freezer with an electronic temperature probe in it. Started at 64 degrees F. Within 1:45 it was at 32 degrees F but totally liquid. After 2:30 it had a thin layer of ice around the inside maybe 1/32" thick that you could easily crack. At around 4:30 it had big chunks of ice in it and the bottle was looking a bit taught, but overall it was still pretty slushy. It's now 6:30 and there is still liquid water in the bottle, the temperature is still 32 degrees F but it's mostly ice. As it froze the freezer came up from 0 degrees to 2 degrees F presumably because of the heat transfer from the state change to solid ice.

I'd say the math works pretty well. If I can find a sleeve of foam core I'll try it again.

9B

How about testing the counter-intuitive experimental finding that warm water freezes quicker than cold (the suggestion is convection currents increase heat transfer).

Mike