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This week-end I build a small solar heater of 1.2meter by 1.2meter. Basically a closed box with the back painted in black and front cover with a transparent plastic. I trying to know the rough performance of it but I get a bit lost on how to calculate it.

The numbers I have:

-The panel is 1.44 square meter ( not sure is required ).
-The fan suck the outdoor air at 10 Celcius in the panel.
-The fan push the air out from the panel at 50 Celcius at noon.
-The rated fan CFM is 16Cubic Feet Minute.
-I don't know the relative air humidity exactly. Let's say 40% but I don't use it for approximate result. I don't want to be so precise.

Here is my wrong calculation:

16 cfm = 0.45 cubic meter minute.
air heat = 721 J/Kg * Celcius
air density = 1.29 Kg per cubic meter
temperature difference = 40Celcius
0.453 cubic meter of air = 0.45 cubic meter x 1.29Kg/cubic meter =  0.58 Kg
721J/ 0.58Kg * 40Celcius = 721J / 23.36 = 30.86J

It's where I get stuck. 30.86Joule is per minute ? So I have to multiply it by 60 to get it in hour ? 1800J ... so 0.5watt/hour? It's not look like correct to me for all that heat come out of my panel. Don't multiply it by 60 give me worst result.

If it's right , my panel have poor performance ! :)

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closed as off-topic by Aaron Stevens, Bob D, John Rennie, Jon Custer, Kyle Kanos Oct 17 at 11:40

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Here's a formula you will need. I'm sure you know it already:

$$P = \frac{\Delta E}{\Delta t}$$

Here's the major problem you have. In your post, I see lots of effort to calculate $\Delta E$, with temperature changes and such. However, calculating how much time the air spends inside your heater is rather difficult because the air goes in and churns around. When air comes out, you can't be real sure how long it spent inside the heater.

I see that your calculations involve cubic feet per minute. You seem to be making an assumption that if, say, five cubic feet per minute is moved out by the fan, then those five cubic feet spent a minute in the heater. You're not completely out of your mind for thinking this, but this is dramatically oversimplified. The cubic feet per minute rating is just a rating - the amount of air moved through the fan may vary wildly from the rating.

I'll allow others to post ideas about how you might estimate the wattage (power) of your heater.

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  • $\begingroup$ Yes , you right. I don't know how much air go in but stay longer , so get hotter , than the air it's coming out. But for me it's don't matter , I just want to know what is coming out of the panel. How much power it's take to heat 16 cubic feet in a minute. I d'ont know if I'm wrong in that's thinking. And yes , the 16CFM is probably not accurate and can vary depends on the payload of the fan ( how easy it can move the air ) but it's the only value I have. $\endgroup$ – Jean-Francois Gallant Oct 16 at 10:41
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Okay today I'm less tired and I look at all that values again and I find a more "closer to be real" answer:

- 16 cubic feet minute = 0.45 cubic meter minute.
- 0.45 cubic meter minute = 27 cubic meter in a hour.
- 27 cubic meter of dry air = 27cm * 1.29density =  34.83Kg.
- Temperature difference stable after a hour = 50C from outlet  - 10C from inlet = 40C.
- It's take 721joule to raise 1Kg of dry air by 1Celcius ( 721J/Kg*Celcius ).
- 34.83Kg * 40C = 1393.2KgCelcius.
- Joule = 721Joule * 1393.2KgCelcius = 1 004 497.2 Joule.
- 1 004 497.2 Joule = 279.02 watts hour.

The only things I don't know is if the fan really push 16CFM on this task but I can't verify it so I assum it's will do it. I let's others members to comments this answer or give another better one.

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