I'm starting a project incorporating many green/efficiency concepts. I’m pretty good at math and have built a computer model with solar calculator, surface (window), and overhang features. In its current form the model takes a nominal solar energy value (from outerspace = about 1.36kW/m2) and corrects for arimass, angle, and overhang. I now want to work in actual window parameter values to calculate actual heat gain/loss. This, of course, will involve the all-important “U” and “SHGC” inputs.

Where I’m stuck right now is that while I have what I think is a pretty good “incident solar energy” value for the window, I don’t exactly know how this value translates to heat (which I’ll eventually put in Btu) for a given window with given solar heat gain coefficient.

I have not spent too much time with google but so far I have not found a clear answer and more importantly I do not have the equations that handle this.

How does incident “solar energy” translate to heat? I assume once this is understood the SHGC is a direct attenuating factor. This should be fairly basic math – does anybody have some good pointers?

Very best,
-Jeff

Btw: the model I have has some nice graphical outputs including a 3-D full day/full year incident energy surface plot. I’ll post a link if there is interest.

Thanks,

I downloaded and played around a little bit with the LBNL Windows5 program. It is very nice and easy to use.

By reading some of the documentation for this program I think found the more direct answer to my question. http://windows.lbl.gov/software/NFRC/NFRCSim5.2-July06Version.pdf with a little more detail in this document: http://scholar.lib.vt.edu/theses/available/etd-04212001-152253/unrestricted/Chapter3.pdf

Question: Assuming manufacturer numbers are real (not sure how big a leap of faith this is) taking “U” and “SHGC” as well as a calculated total incident irradiance from the sun (this can be found a number of ways and I already created those models) what is the instantaneous solar heat gain (W/area or Btu/area). This value can then be integrated over the solar day (where incident energy, angle, and shading all vary).

From previous research:
Incident angle – need the sun’s position and surface (window) orientation
Incident Energy – need solar constant and attenuate for air mass, angle, and overhang (shadow).

From Chapter 3 of the Window5 documents:
q = Ut*Apf(tout-tin)+(SHGC*Apf*Et)

Where: q = instantaneous enery flow (W)
Ut = overall coefficient of heat transfer (W/m2-K)
tin = interior air temp (K)
tout = exteriour air temp (K)
Apf = total projected area of fenestration (m2)
SHGC = overall solar heat gain coefficient (unitless)
Et = incident total irradiance (W/m2-K)

I did not try to balance the equation but in one place the irradiance number is in W per meter squared degrees Kelvin. I’m not sure where the degree Kelvin comes from… Thoughts on this usage?

Cheers,
-Jeff