# Calculate heat transfer out of a basement [closed]

I am thinking it should be possible to calculate heat loss out of basement based on the temperature of the walls, but am not sure how to do it (what the equations are).

For example, lets say the measurements are as follows:

• ambient air temperature in basement (same as ceiling): 10 degrees C
• temperature of windows (10 square feet of window): -8 degrees C
• temperature of upper wall (250 square feet of upper wall): 1 degrees C
• temperature of lower wall (720 square feet of lower wall): 4 degrees C
• temperature of floor (900 square feet of floor): 6 degrees C

Assume the ambient air temperature is maintained at 10 degrees C.

What is the rate of heat loss through the walls, windows and floor of the basement?

Thanks for helping me figure out the right method to calculate this.

## closed as off-topic by Brandon Enright, Ali, David Z♦Oct 29 '14 at 20:27

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• I think you would need more information. Either how much energy per time interval is needed to maintain that room temperature, which would then already be your answer, or how thick your walls and windows are, what the material is exactly and what the temperature outside is. – Jupith Aug 25 '14 at 21:26
• Why does your basement have windows? – akrasia Aug 26 '14 at 15:50
• You would be better off with dedicated building energy efficiency websites. They will include information about the average ground temperature, the performance of typical building materials and so on, and will do some of the calculations for you. They will also use the same units to describe the building materials as the ones you will find on the label. – akrasia Aug 26 '14 at 15:54
• That does not make sense to me. My understanding is that the temperature difference between two layers implies the heat transfer rate between the layers. For example, lets imagine that the transfer rate was infinite, then obviously everything would be 10 C. Are you a physicist? – Ambrose Swasey Aug 26 '14 at 16:02
• No I am not. Anyhow: You have a temperature gradient inside the room as well as in the walls and windows. Ideally, if you have a material with area A and thickness L that is in contact with two different heat reservoirs T1 and T2, the heat transfer rate is proportional to the temperature difference multiplied with the area of your material and divided through its thickness. So depending on the position between the two reservoirs different parts of the material have different temperatures. What do you mean exactly when you talk about the temperature of your walls? – Jupith Aug 26 '14 at 19:43

After studying this problem, I have found a method of solving it. First of all, since the ceiling is the heat source, we don't know what it needs to be to keep the room at 10 degrees C. Obviously it will have to be hotter than 10C. Secondly, the temperature will differ in basement. For example, it will be hotter near the ceiling and colder near the floor. Therefore, let us assume as a goal to keep the center of the room at 10 degrees C.

For simplicity let's ignore convection and radiation. Since the heat source is at the top of the room convection will be much less than if it were at the bottom anyway.

We can estimate the heat flux by taking as an average the distance of each surface to the center of the room. From the square footages listed it appears the room is 30 x 30 x 8 feet high. Therefore a wall is 30 x 8. The shortest distance from each wall to the center is 15 feet. The shortest distance from the ceiling and floor to the center is 4 feet. The average distances can be computed using Mathematica as follows:

NIntegrate[
EuclideanDistance[{x, y, 0}, {0, 0, 15}], {x, -15, 15}, {y, -4,
4}]/(30 8) = 17.3723 ft (or 5.295 meters)

NIntegrate[
EuclideanDistance[{x, y, 0}, {0, 0, 4}], {x, -15, 15}, {y, -15,
15}]/(30 30) = 12.2824 (or 3.744 meters)


We can now estimate the conductive heat flux using Fourier's Law q = k * area * deltaT / distance, where K is the thermal conductivity of air which is 0.024 W/(m.C).

upper wall heat flux: 0.024 * 23 * (10 - 1) / 5.3 = 0.9374
lower wall heat flux: 0.024 * 67 * (10 - 4) / 5.3 = 1.82
floor heat flux: 0.024 * 84 * (10 - 6) / 3.7 = 2.179
window heat flux: 0.024 * 1 * (10 - -8) / 5.3 = 0.0815


This totals to approximately 5 watts.

We can also compute the temperature the floor will need to be to maintain the center temperature at 10 degrees C:

0.024 * 84 * (T - 10) / 3.7 = 5


Therefore T is approximately 19 degrees celsius.