Solar heated Water Tank, Room heater project?


I have a fun idea I want to build. For a off grid location bungalow type room.

The idea is to use some black coiled hose encased in Perspex as solar hot water collector.

Which a pump will cycle the water through the coiled hoses and a unknown required sized Liter thermally shielded container.

And to use an old small radiator of some kind to bring that heated water into a heat source for the very very cold nights.

So I have some questions. I hope a thermal engineer might answer.

The space to heat up Approx a square space of (3m wide, 6m long 2.5m high)

I want to change the temperature in that room from 10 degreesC to 25 Degree C

The water when cycled through the radiator would be a toasty 65degree C.

I want to know the amount of water in liters that would be required to provide that kind of heat exchange.

I am sure I can add additional thermal protection to the bungalow to attempt to create a scenario where I heat it once and let it stay warm for a longer period of time.

This idea of water thermal heater has interested me for a long time.

Anyways thanks for reading

3 Answers

  • 1 month ago

    Full of this device and plan can be found in youtube video. Some use low cost and no cost material to make this water heating with direct sun shine, like soft drink bottle cover with black garbage bag.

  • 1 month ago

    the energy required to heat the room is a complicated calculation. The volume of air to be heated is minor compared to the objects in the room, and the heat loss in walls, doors, windows. It also depends on the location and size of the radiator, and it's surface temperature. 

    "I want to know the amount of water in liters that would be required to provide that kind of heat exchange"

    the amount of water is irrelevant, what is important is the water flow, if you want to maintain the temperature. Flow is in liters per minute or gallons per hour, or some such. 

    If I had to guess, given the size of the room, assuming well insulated, and with typical heat losses, assuming outside temp of 10ºC, I'd say about 5000 BTU/hr or 1500 watts, which is 1500 joules/second.

    1 liter of water (1 kg) at 65ºC, relative to 25ºC has an energy content of 

    E = 4 kJ/kgC x 1 kg x 40C = 160 kJ

    so to get 1500 J/s, 1.5 kJ/s, you need a flow of

    flow = 1.5 kJ/s / 160 kJ/kg = 0.01 kg/s

    or 0.6 kg/min

    if I didn't make a arith. mistake....

  • 1 month ago

    The major problem here is in energy storage.  You need the heat at night and the water in the system would only heat in the daytime and cool very rapidly.  You might also have to worry about using antifreeze to keep the water from freezing in the hose and splitting it. 

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