We want to ensure that the electronics we send up continue to function in the cold air we hope to encounter. Temperatures vary by day and by area, but this data from the NASA Pathfinder is a reasonable estimate of what we'll encounter.
| Altitude (ft) | Temperature |
| 10,000 | 4c / 40f |
| 20,000 | 6c / 21f |
| 25,000 | -16c / 3f |
| 30,000 | -28c / -18f |
| 35,000 | -40c / -40f |
| 40,000 | -50c / -58f |
| 45,000 | -63c / -81f |
The various bits of electronics we're using vary in their tolerance of cold, so we're going to be conservative and try to keep all of the electronics above freezing temperature. In steady state this means having to maintain a temperature differential of around 75c or 110f. This involves two things: 1) providing some insulation for the electronics and 2) providing a heat source to replace the heat that dissipates through the insulation.
Insulating the electronics
For a camera, embedded computer, servo, GPS, modem and batteries, our guess is that we needed a cavity of around 3" x 5" x 7". This is around .07 cubit feet and has a surface areas of .99 square feet. In the interest of cheap and easy, we decided we should make our insulation cavity out of polystyrene, or good old fashioned Styrofoam. Obviously the thicker we make the walls of the cavity, the less heat we'll lose and need to put back. Expanded polystyrene has an insulating quality (or R-value) of 3.85. (Wow, this chart says "Phenolic Foam" has an R-value of 8.3, I wonder what it is and where we get some).
How much heat will we need to add?
The formula for the hourly expected heat loss is:
Q = U·A·T
Where:
| Symbol | Meaning |
Units |
| Q | Total hourly rate of heat loss | (Btu/hr) |
| U | Heat transfer coefficient (The reciprocal of R-value) | (Btu/hr-ft2-°F) |
| A | Net area for heat transfer | (ft2) |
| T | Temperature difference to be maintained | (°F) |
So, with .99 square feet, an R-value of 3.85 per inch and a temperate difference of 110 degrees Fahrenheit, we get
Btu/hr = 1/ (3.85 * inches of polystyrene) * .99 * 110 = 28.2 / inches of polystyrene
The conversion to watts is (Btu/hour x 0.293) so
Watts = 8.3 / inches of polystyrene
This means with an 8 watt appliance (oh say a light bulb) in our 1 inch thick polystyrene box, we could hang out at -80 degrees Fahrenheit until we ran out of power and the electronics should be happy.
| For a sanity check, we built a box of 1"
polystyrene the appropriate volume and put inside 750 ml of water and a
wireless thermometer. The water started out an initial temperature of 98
degrees Fahrenheit and the ambient freezer temperature was 5 degrees
Fahrenheit. Initially, with a temperature difference of 93 degrees, we'd
expect to dissipate heat at a rate of 1 / 3.85 * .99 * 93 = 24 Btu/hr.
At .004 Btu per calories, we can compute the rate we expect the temperate of
the water to fall by. (For the first minute we'd expect the water to fall 24
btu/hr / 60 mins/hr / 0.004 Btu/calorie / 750 g water * 1.8 f/c = .24
dergees Fahrenheit). Since the thermometer only reports in every couple
minutes and since the box was taped together using packing material from a
coffee maker, we're not looking for exactness, but it would be nice to be
within a factor of 2 of what the science says. Shockingly the predicted and measured numbers turned out scarily similar as shown below. If this was for an assignment in school, I would definitely add some error to my measured data some for fear of being accused of cheating. So it look like the "8.3 / inches of polystyrene" is something reasonable to bank on.
|
|
 |
|
Another approach to keeping the electronics warm
Rather than using a battery and a heating element to keep the cavity warm, another approach would be to include a substance like water that sheds a considerable amount of energy before going into phase change and freezing. This would not keep the cavity from quickly getting to 32 degrees, but it would keep the cavity at 32 degrees for as long as it took for the water to freeze. The amount of energy requires to change 1 gram of water from ice cold water to ice is 80 calories. Assuming the water was warm (100 degrees Fahrenheit) when placed in the cavity, it would take 3340 calories to freeze 1 ounce of water. Our heat dissipation rate of 28.2 Btu/hr is equivelant to 7100 calories/hr. This translates to an ounce of water taking 28 minutes to freeze in a 1" thick polystyrene box at 110 degrees of temperature difference. This means that in lieu of an active heating solution, we could instead build the cavity from 2" thick foam and by including 6 ounces of warm water in the box, we would be able stay for 5.6 hours at -80 f without our electronics getting too cold. Cool!
