Cold Safe Room
 

I am wondering if anyone with a science background can help show me how to calculate heat loss.

The concept is to build an insulated room inside a house that is small enough to maintain a temperature heated by body heat.

For example, if I had a cabin heated by a woodstove, but I got a fever and could not keep the woodstove going, I could retreat to a cold safe room that is insulated well enough that body heat alone would keep it at a certain temperature.

If I look at building materials, such as this 2" foam board, it is rated at R 7.7 .

If I built a room of this material in a size of 8' x 8' x 8', and the external temperature was 20F, how warm would it be inside the room if heated by body heat?

Assume that there is a perfect heat exchanger providing fresh air.

I am mainly interested in how to calculate this, what units, etc.

The R-value is a measure of specific thermal resistance, and R10 = 10 ft²*°F*hr/(BTU*inch). I much prefer the metric meters*Kelvin/Watt. The USCS (not metric) system discourages good practices by making it harder to check your units during a calculation.

The equation for heat flux is:

Q = ( surface area * deltaT ) / ( R-value )

If you have an 8' cube, its surface area is 384ft². Assume 9" thick of fiberglass batting at R4 per inch, giving R36, plus two surfaces of a wall with R 0.6 each.

A healthy human at 2000 Calories/day puts out 330BTU/hr, so we'll set heat flux equal to that.

330BTU/hr = ( 384ft² * deltaT ) / 37.2 ft²*°F*hr/(BTU)

deltaT = 32°F, and the body heat of a healthy person will hold a 32°F temperature differential across the R37.2 barrier. What will the temperature be on the other side of the wall?



This mathematical model of a room is too simple to yield anything but a first approximation. If any of the room's six walls are exterior walls or floors, or if sunlight is incident up on them, the math changes. There are also transient effects to consider: the room will take a long time to reach its steady-state temperature.

(Building Heat Loss in BTU's per hour) =
[(Building Total Surface Area in sq.ft.) / (Surface Area "R" value)] x (Temperature Difference)

Size of one wall = 8 X 8 = 64 ft sq.

Area of room = 64 X 6 (including ceiling and floor) = 384 ft sq.

Assume you start with a room temperature of 72 degrees.

Heat loss per hour = (384/7.7) X (72-20) = 2500 btu/hr.

A human on a 2000 cal/day diet gives off about 330 btu/hr as heat. Thermal Efficiency of a Human Being

2500/330 = 7.56. So we would need to increase the R value to 7.56 times the current value to maintain 72 degrees in the room with only one person in it. Or decrease the surface area by 7.56 times. Or increase the number of persons to 7.56 times (add 6.56 people to the room).

Figuring it another way, we can rearrange the equation to determine the temperature that will be maintained with one person in the room: (Temperature Difference) = (Building Heat Loss in BTU's per hour)/[(Building Total Surface Area in sq.ft.) / (Surface Area "R" value)].

Substituting our known values we have: (Temperature Difference) = 330/(384/7.7) = 330/49.9 = 6.62 degrees. So the body heat of one person will maintain a 6.62 degree temperature difference in the room. 20 + 6.62 = 26.62 degrees in the room with a 20 degree outside temperature.

It is interesting that surface areais so important. One way to make a cold saferoom would be to get two sturdy dome tents, one about 4 foot less in diameter than the other. Set up the smaller tent inside the larger tent, and fill the space between with insulation, such as celluose insulation. Beef up the inner tent structure with additional fiberglass rods or pvc pipe. Put the tent on an insulated platform and access it through a trap door on the bottom. Instant Igloo!

In any case, both for thermal efficiency and material cost efficiency, a dome shaped structure is required for extreme insulated structures. At least, as long as you can cut the pieces of the dome out of the material without a lot of waste.

Yep, Eskimos weren't stupid, and the igloo has the correct shape.

Patrick and I have made the "one dimensional heat transfer" assumption, where heat flows only perpendicular to the surface. That's fine for rectangles and objects that are much larger than they are thick, but for the igloo you describe, you may want to break out some semi-empirical correlations for spheres or cylinders. Interesting fact: adding additional insulation to a cylinder can INCREASE heat transfer by increasing the outer surface area faster than it increases the insulation thickness. I don't believe this applies to any geometry but cylinders.

I think this is one instance where it sounds good in theory, but . . .

It will be a LOT easier and probably cheaper to build a structure in a rectangle or cube shape that will use readily available insulating materials and add more insulation than to make a dome. You could make a superinsulated wall with 2x6 framing or offset 2x4s, put fiberglass inside the wall, then add foam on both sides. Last time I looked at geodesic domes they had to be specifically engineered and used special attachment plates at each joint.

Also, the usable volume inside a rectangle will probably be much greater than in a dome. I'm not really fond of bumping my head on the wall every time I move around a room.

Let's look at the example you presented in your first post and compare a dome to a cube structure. For a dome with an 8 foot ceiling (in the center), the radius of the dome would be 8 feet.

Area of sphere = 4PiR*R

Area of dome = 4PiR*R/2 = 4 x 3.14159 x 8 x 8 / 2 = 402 ft sq.

Area of floor = PiR*R = 3.14159 x 8 x 8 = 201 sq ft.

Add area of floor to area of dome: 201 + 402 = 603 sq ft of area for total structure.

603/384 = 1.57 times as much surface area for the dome.

So a dome you could "live in" would be worse, from an energy standpoint, than a cube. Granted, you get more floor area with the dome, but you'd be hunching over much of the time to keep from bumping your head on the ceiling/wall.

If you set the surface areas equal on both the dome and cube, the heat loss would be the same, so there would be no advantage thermally to either one. But the ceiling height will be lower for the dome, perhaps too low to make living in it acceptable. And if you want to put a door in the dome, you will have to flatten one of the walls and have the dome at least as tall as the height of the door, so the amount you can lower the ceiling will be limited. The dome will be harder to build, but it would offer more floor area to store survival equipment or to put thermal mass.

Sleeping bag?

And small cylinders at that. In college, we had a lab that centered on this idea, and found that in theory, this is true, but in practice, it is only true on hypodermic-needle-size cylinders, and not true for larger cylinders. So feel feel free to pack on the insulation!

If you are actually going to attempt something like this, you will want to make sure things are sealed up tight, and then have a heat exchanger for getting fresh air. If there is much of a draft, you will find that you lose most of your heat just to that. If you have it all sealed up with no fresh air able to get in, you could suffocate.

It sounds like the most economical way to do this would be to take an existing room and build an inner wall frame with ceiling and raised floor, and fill the wall space with insulation.

I don't plan on doing this, I am mainly interested in how to calculate it. I am interested in how a small, well insulated house could be built with minimal heating needs, and how solar heat can be stored to reduce fuel use.

You could make a structure with a Trombe wall and high thermal mass:

Trombe wall - Wikipedia, the free encyclopedia

Thermal mass - Wikipedia, the free encyclopedia