All very well, a solar heating, but what does it yield? 50° hot air is of course nice, but is it significant? A calculation.
The system has a fan that moves 70 m³/hr of air, in principle.
The heat capacity of air is 1,005 kJ / m³ it therefore costs 1005 J of energy to make one kg of air one degree warmer.
1 m³ air weighs 1,225 kg.
The outside temperature is about 15º, while 50º is output, a temperature increase of 35º.
If that happens during one hour, there is 70 m³ air 35º warmer.
That costs 70 m³ * 1,225 kg * 1005 J * 35 = 3016 kJ in one hour.
Watt = J/second, so the capacity is 3016 kJ / 3600 s = 834 Watt
The sun emits around 2.7 kW of energy per m² this thing has a surface area of 2.1 m ², and thus captures 2.1 * 2.7 = 5.67 kW energy.
This means that the yield is about 834 / 5670 * 100 = 15%, which is not very different from the efficiency of electrical solar panels (15 - 20%)!
Is the capacity significant? An electric radiator has a capacity of 800W (low) to 1400W (high), the kitchen has a volume of 140m³
I have materials for three solar heaters, so perhaps a few degrees? I hope that will make the difference. And it's free, that makes quite a difference.
You can view solar heaters as a "remote window". Normally the sun shines through windows, and heats up an interior. However, this house hardly has windows, so we'll use these "remote windows" instead.
You can wonder if the 70 m³ per hour air really is achieved is a good question. Perhaps next time.
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