Posted by nicksanspam on October 10, 2004, 5:29 pm
Consider a motorized vent actuator instead. How many cfm do you want
to move? Two A ft^2 vents with an H' height difference and a dT temp
diff will move 16.6Asqrt(HdT) cfm or about 16.6Asqrt(H)dT^1.5 Btu/h.
For example, a 14'x96' greenhouse covered with 80% shadecloth in full
sun (250 Btu/h-ft^2) collecting 0.2x0.9x250x14x96 = 60.5K Btu/h with
a 1'x96' opening at the ridge 7' above 2 0.5'x96' slots near the ground
might move 16.6x96sqrt(7)dT^1.5 = 60.5K Btu, so dT = 14.3^0.666 = 5.9 F
above the outdoor temp, or less, counting evaporation from plants and
conductive heat loss through the glazing.
Posted by Mike C. on October 10, 2004, 10:02 pm
Love the math! I am an engineer at Caterpillar.
I do not think that vents will work in this case. It is a small 12X12
pit greenhouse. I have insulated the north side to use it in the
winter time for fall/spring veggies. I would like to use it in the
summer time but I run 120-130 in full sun. I need to keep it at 90.
I have a 12" fan with cross vents and it does a good job but runs full
Anyway I want to paly with solar power! Remember I am an engineer!:-)
Posted by nicksanspam on October 10, 2004, 11:00 pm
At 80 F outdoors, 0.2x0.9x250x12x12 = 6480 Btu/h = 16.6Asqrt(8)(90-80)^1.5
makes A = 4.36 ft^2, eg a 2'x2' gable end vent.
Posted by Bernd Felsche on October 11, 2004, 1:45 am
firstname.lastname@example.org (Mike C.) writes:
So you should know that converting a motor from AC to DC is going to
be very expensive in terms of time and probably materials...
Depending on the current draw of the fan, you're probably better off
with either an inverter to produce the requisite AC voltage/frequency
for the fan motor, or replacing the AC fan/motor with a DC motor.
Cheap DC fans, for moving large volumes of air, can be found in the
automotive industry. Presently they limit you to 12 or 24 volts
nominal unless you can find a competent rewinder to increase it to
42V nominal... you don't get the inverter losses, but need heavier
wire between the fan and electrical storage.
Some electrical storage is probably required so that you don't run
the motor at under-voltage by operating directly from
photovoltaics... the maximum power point trackers you should use
(else your 12% efficiency cells become 8% efficiency on a good day)
also require a "load" into which to drop the current.
The energy storage doesn't have to be overly capacious; enough to
run the fan(s) for at most an hour. Unfortunately, common lead-acid
batteries are not very good in terms of either recovery efficiency
(about 70 to 80%) or their useful capacity which is also only
around 60% of rated.
You can probably install *static* external shade panels that don't
significantly diminish winter insolation, yet prevent the hottest
part of the summer sun from hitting the surface of the greenhouse.
In summer, the sun is between 20 and 45 degrees higher in the sky
than in winter/spring/autumn. Shade makes an enormous amount of
difference in terms of greenhouse insolation.
Do a google search to find any of the hundreds of references to
calculate solar angles at various times of the year at your location
and work out the length and angles for the shade panels. Once
they're in place, they consume no power.
Direct thermo-mechanical use of symptom to solve the problem is an
obvious approach. A black-panel collector with working fluid can
develop a significant thermo-syphon (water can be *boiling* in
summer). The resulting flow could drive an impeller to turn a fan,
perhaps not very rapidly without gearing up, but you can make up for
it in diameter and depth. Probably over-kill for your tiny
/"\ Bernd Felsche - Innovative Reckoning, Perth, Western Australia
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Posted by Toby Anderson on October 11, 2004, 7:44 pm
In this newsgroup (alt.solar.thermal), there is a 'nuance' between
'solar power' and 'photovoltaic' power. The later uses PV panels, the
former might use them, but concentrates on non-electrical methods. You
might try the Photovoltaic group to answer your 'motor' question.
In this 'solar thermal' group, if you want to create airflow, you
would use the principle that heat rises. I realize, that you have a
'pit' greenhouse, but what I don't get is how high the ridge of your
roof is above the ground. How high is it? Being a pit, it's probably
not very high above ground level, so the distance is not that great
and you will not get that much air flow.
I suggest you build a solar chimney on the top of and at one end of
your greenhouse, say 8ft tall (H=8'), by 8ft wide by 2 foot thick.
Have a 1 foot by 8 ft opening at the top and at the bottom (A =
8sqft). Fasten a black shadeclothe to the back masonite board, and put
clear polycarbonate around the other 3 sides. At the other end of your
greenhouse, place a 8sqft vent to the outside. Place misters there as
well. It's likely that the mist will generate some evaporative
cooling. The air in the solar chimney should be hotter than your
130F, but let's assume that it is 130F in the chimney and 100F outside
(How hot is the ambient outside temperature), hence, assuming that
deltaT is 130-100 = 30F. An estimate of your airflow will be:
= 2060 cfm
This is the amount of air which a small to Medium swamp cooler might
Keep in mind, that if the outside air is really 100F, then, unless you
use 'misters', the 'coolest' your greenhouse will get is 100F.
My guess, is that even if you use misters, you will have a hard time
keeping it cool enuf.
I suggest, you consider a 'water window' for your roof to trap alot of
the IR heat, but that's another topic.