fin/wing tanks freezing

I don't recall a thermometer in the tail tank as a requirement in either of
my LS-6s. Maybe because they were early models...

My LAK-17a Pilot's Manual requires an OAT gauge if water ballast is carried
and states further:

Warning:
Flight with water ballast must be conducted at an OAT greater than 1° C
(34°F). If there is a risk of freezing temperatures, all water must be
dumped before freezing temperatures are reached.

Being somewhat of a "test pilot", I often fly with water at below freezing
temperatures (it's cold at 18,000' MSL regardless of seasons), but I
generally don't spend more than a couple of hours at those altitudes and
temperatures. I also spend plenty of time on the descent to allow things to
warm up at a reasonable rate.

Having said that, I've had a stuck wing dump valve more than once. It's not
noticeable until the glider comes to rest.

"Andy" wrote in message
...
On Jun 6, 3:37 pm, wrote:
What I was wondering is whether there are any gliders out there with
thermometers in the tanks.

I think some, if not all, LS gliders, have tail tank thermometers and
that it was a requirement of the type certificate. Never understood
why it should be required for one make/model but not for gliders from
other manufacturers.

When I had my ASW-19b I was concerned out wing ballast temperature.
Since the 19 uses bags it was easy to fit a thermistor probe under one
of the ballast bags and connect to a cockpit thermometer. (Radio Shack
indoor/outdoor thermometer with a connector break in the outdoor probe
lead). I was surprised at how warm the ballast stayed despite long
cold soaks at altitude.

The risk of tail ballast failing to dump is taken seriously by some
manufacturers but not by others. For the ASW-28 the aft cg limit
moves forward as mass is increased. If I operate within limits I
will not exceed the aft CG limit if my tail tank fails to dump.

On the other hand, the Duo Discus handbook recommends an increase in
minimum front seat mass if a tail tank is fitted. It then goes on to
say it's only a recommendation and you can ignore that safety
protection if you want to.

Failure to dump the tail ballast is not just a freeze risk. The valve
can jam or the cable can break when it's nice and warm outside. I've
even seen a well meaning crew tape over ballast vent holes.

Andy

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

After reading this post I am just glad my elementary school math teacher was wrong when she said I needed to learn math because I wouldn't always be walking around with a calculator.

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