Calculate heat transfer out of a basement 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.
 A: 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.
