Tue Nov 02, 2010 11:47 am by Fish4Fun
Ben,
Unless I am missing something, I think you may be. Heat transfer is a tricky business, so it may well be me who is missing something, but....
1) Isn't the amount of energy in the sunlight measured in W/m^2, and therefore fixed by a combination of available energy, area of the window and emissivity of the glass?
2) Understanding that the sunlight is made up of a wide spectrum of electromagnetic energy, doesn't the vast majority of that energy end up as low-grade heat regardless of what surface it strikes? Even highly reflective surfaces absorb some energy, unless the majority of the light energy is simply "passing through the room", then the room will eventually (micro seconds) absorb nearly 100% of the energy in the form of low grade heat regardless what you do.
3) I agree that placing a black object in the window will concentrate the heat gain for the light striking the black surface, and that the black object will experience a sizable heat gain compared to the more reflective objects around it, but I would argue that the overall heat gain in the room is un-affected.
Now, in the case of a "black box" placed outside with an air exchange into the room, in this case the amount of heat gain to the room involves the area of the black box and your ability to maximize thermal capture and minimize thermal losses. This is a fundamentally different case than trying to use existing windows to capture heat that has already entered the room.
I am NOT stating your theory is wrong, I am simply questioning the theory. I would love to see an experiment. I would think that a 1m^3 insulated box with a glass pane X cm^2 would do nicely. "X" should be proportional to the room volume/window area. That is to say if your room is 6m x 6m x 2.7m, and your total sunward facing window area is 2m x 2m, then 4/97.2 = .0411 would translate to 411cm^2/M^3, or roughly a 20.25cm x 20.25cm glass pane in a 1m x 1m x 1m insulated box. Build two boxes and set them side by side with the same size window and place thermometers in each. Paint the inside of one box white and the inside of the other black. Set them both facing South and record the temperatures inside each box. I would expect that as the area of the glass increases with respect to the South facing "wall" area, that the heat gain in the "black box" increases with respect to the white box. I would also expect that as the "depth" of the boxes increases with respect to the glass area that the difference in temp rises decreases. As a final experiment, you might paint the inside of both boxes white, and add your "black strip". I would again expect a significant heat rise advantage for the black strip" only when the the area of the glass exceeds some percentage of the South facing wall.
To scale this experiment to an affordable size, one might use a pair of small Styrofoam coolers and some microscope slides. Or you might insulate a pair of shoe boxes.
I would love to know the real-world results :-)
Fish