solar air heater calculations

If you don't like math or physics, then you do not want to watch this. If anyone actually enjoys watching this...god bless you...my speech was jerking around too much to be enjoyable. This is torture even for me to listen to. The equations are:
Watts lost due to blackbody radiation per meter squared =

0.95*5.67E-8*[(kelvins black surface)^4 - (kelvins outside)^4]

It's not this simple and it's not this much heat lost if the front face is glass. Glass and thick plastics block basically all far infrared heat at these temps. But this in turn heats the glass which still causes heat loss, especially since the R factor on the outside surface of the glass in winter is about 0.3 (not 0.17 as stated in wikipedia). The outside ambient is not really the temperature to use, because the sky is not a solid object and space is at near absolute zero. It's actually a substantial difference, about 20 C (20 K) lower, or more, giving a larger heat loss. A paper said the following is a good substitute for the ambient term: 5.31e-13*(kelvins outside)^6.

Also remember the sky is reflecting some light giving more energy input than the amount directly from the sun.

sq ft = 10.75*meters^2

Primary heat loss Watts = (F avg inside - F outside)*(sq ft)/3.41*(1/Rfront + 1/Rrear)

This ignores the additional heat lost through the sides. If you have air flow against the front surface like my design and it is single panel, then R in the front might be 0.5 instead of the 0.75 in the video, which is a big loss. Also, my R=4 for the back should have been R=3, and I should have used the temps of the backing, not the average inside air temp. So really, my math should have been:

(98-71)*32/3.41*1/0.5 + (140-71)*32/3.41*1/3 = 772 Watts instead of 401.

R factors usually include the effects of black body radiation which is one reason winter R factors for building materials are not as good as summer R factors. But since the temps in the boxes are getting kind of hot, it's good to get an understanding of black body radiation even if there is some error in how I'm doing these calculations. I have not found a reference that really calculates the total heat loss correctly because of this mixing up of black body radiation and R factors. One way might be to use traditional R factors, then substract block body losses that you calculate at normal house temps, then add in the black body temps you calculate for the box. It would be nicer just to have R factors that do not include black body radiation. The only way to really know is to measure heat in and heat out yourself.

Measured heat output of any design = Watts = CFM*(F out - F in)/2.94
This assumes the CFM was measured at 68 F. You measure the CFM by filling up 5 sealed trash bags like I did (CFM ratings are unreliable) and timing it. It's best to fill up the trash bags with indoor air which is near 68 F and then see how long it takes to empty the bags by attaching them to the inlet when under normal full operating conditions. If you measure hot outlet air, the results need to be multiplied by (460+68)/(460+F out) which lowers the results in proportion to the amount that the outlet temp is above the 68 F in absolute temperature terms (Rankins = 460+F)

Do not use low E glass that they use in well-insulated windows to block the black body radiation because those coatings also reduce the amount of light that can get through (like 70% transmittance for 2 panes instead of 85%)

Heliostats which reflect light directly into the house (bay window or sliding glass doors) are harder to build due to having to track the sun and the mechanics. But they are about twice as efficient and the materials can be 10 times cheaper. I am working to get a design completed for 2012-2013 winter. See my 25 second video. Two of those cut my heating bill in half this winter from $200 to $100 per month and we stay warmer. Cost of materials was $100, but I don't have the electronic tracker yet: I adjusted it every hour by hand! For 3 months so far! Every sunny day!

Category:

Science & Technology

Tags:

• solar
• air
• heater
• calculations

[HudsonKnowsEconomics]

I finally figured out the black body calculations when there is a glass front and it is a bit complicated (I could not find a reference, but it's a differential equation). The end result is that the losses will not be as much as I've indicated because the glass is blocking all the radiation heat from immediately escaping, but it matters on cold days, especially you let the black part get hot. The primary goal for efficiency in the design of these things is high air flow to not let it get hot.

[HudsonKnowsEconomics]

@Yaheuben66 the unpainted side feels hotter to touch because without the paint as a thermal insulator, the metal is transferring heat to your skin better. I don't think the unpainted side will heat flowing air better than the painted side, especially because with the higher air flow it is not supposed to get too hot. I like the black plastic trash bag method due to being cheap and light. A problem is that you can't let the air flow go to zero while exposed to sun or it melts.

[rlidrazzah]

Thank You!!! The equation helps me judge my design so much easier.

[Yaheuben66]

WoW!! Thank you, thank you, thank you for putting this information out.. I have been bouncing this problem around in for a year now... Great information!

I have thought about the "collector"-- I thought that aluminum foil (painted black with a "naked" back) placed mid-channel would be best-- aluminum is a good heat sink/shedder

I have a piece of aluminum plate painted black-- the back side gets WAYYYYY hotter than the black-painted front (tested by conduction/touch, anyway) in full sun...