Mary Hinge wrote:
Ian Stirling wrote in message ...

You'r looking at a solar gain of up to around 10Kw.
(need more information about total glass area in the sun direction.
for more accuracy)
Air absorbs around 1KJ/m^3/C.
So, to keep the temperature rise down to 10C, you'r looking at
.1m^3/s of airflow.
To keep it within a couple of degres, more like half a cubic meter
a second.

A 30cm fan should easily do this, requiring only an output speed of
1-5m/s or so.

Very interested in this thread - in a similar position with the house
we've just moved into recently. This time a large lean-to conservatory
- 5m wide by 3m deep, east facing against a two-storey house so mostly
in shade in the afternoon. Solid walls to south and north sides, fully
glazed to the east. Roof is very shallow pitch triple-wall
polycarbonate.

I would prefer to avoid air con due to high up-front and running
costs. Conservatory blinds seem to cost an awful lot as well. What
sort of air extraction rate might be required in this case? Any chance
one of those 275m3/hr Xpelair jobs might help, or are we talking major
air-shifting?

For ballpark use, the total heat gain is 1Kw/m^2.
This area is measured at right angles to the sun.
As the day goes on, the area rises and falls depending on the angle
which the sun is shining onto it.

I'll assume that the 5m is against the house, and that it's 2.5m tall.
In the morning, you've got some 5*2.5 = 12.5Kw.
As a very rough ballpark, 1Kw raises the temperature of a 1m^3
by 1C in 1 second.

275m^3/h is (/3600) =.08m^3/s.
Heated by 20C, this will absorb .08*20 = 1.6Kw, which is about an eighth
of what you need.

So, if this was the only means of cooling, it'd heat to some 160C.
The polypropylene will probably melt, and let the heat out first
(or more accurately, it'll simply go out through the walls.