> > David WE Roberts <nos...@btinternet.com> wrote:

> >> > I have in mind a wooden frame, containing a long folded tube behind a
> >> > double-glazing panel and backed with a reflective surface.

> >> > The tube would be a long series of aluminium drinks cans, painted matt
> >> > black.

> >> > The will have ducts to the garage; one to draw cold air out of it, and
> >> > the other to bring warm air from the heater back in.

> >> > A small solar-powered fan would draw air through the tube.

> >> I assume that, given that hot air rises, you will have a strong enough
> >> fan
> >> to drive the hot air down to floor level before exiting the ducts?

> > How powerful would a fan need to be?

> I was hoping you could tell me ;-)

> Firstly it would depend on the volume of your garage.

> {Someone throw me a mathematician and/or engineer quickly}

> If your garage is at 5C and the air in your heating array is at 35C then you
> have a 30C temperature difference.
> If you want your garage to be a nice pleasant 20C then you have to replace
> half the volume of your garage with warmed air to achieve this - but quickly
> enough to compensate for the heat loss through the structure.
> Hmmm.....half the volume of your garage every 30 minutes to allow you to get
> up to temperature in 30 minutes (which does not seem an unreasonable taget)?
> So work the volume of your garage out in cubic feet, halve it, divide by 30
> and you know how many cubic feet per minute you have to shift.
> This specifies your fan(s).
> These specify your solar panel.
> You would have to allow a bit more push because you are having to push warm
> air downwards whereas I think most extractor fans expect to be pushing air
> horizontally.
> The hotter your heat exchanger gets, the less volume you should have to
> exchange.
> e.g. get your heat exchanger to 65C and you only need to exchange 25% of the
> garage volume of air.

> Someone who has already done this or at least the calculations should be
> along shortly (I hope).

> Cheers

> Dave R