Posted by Mark & Shauna on May 5, 2004, 8:28 pm
Hello to the group,
I am wondering if some of the more technically savvy people out there
could speak to what kind of air flow one could expect to achieve using a
solar chimney of sorts in place of large fans in a greenhouse?
The situation is, we are off grid and have a couple decent size
greenhouses which to date have been using simple vents for cooling and
air flow. However, we are not getting enough flow which is causing some
problems. We have actuated vents in the gables at the peak which open at
a set temperature now but even when full open the air in the greenhouse
is not moving much on a calm day.
There is no way we can build another power system to drive large enough
fans so I thought of the solar chimney idea. Perhaps an 18"x18" I.D.
chimney of a given height, black on the outside, etc.
I am wondering what kind of flow we could achieve on an average sunny
day via this method. Any guesses? Would it be something you could feel?
Is it a wild goose chase?
Thanks for any input,
Mark
Posted by Mikemotorbike on May 6, 2004, 2:09 am
I have read that a solar chimney moves the same as a 1/4 hp motor.
-mikemotorbike
> Hello to the group,
> I am wondering if some of the more technically savvy people out there
> could speak to what kind of air flow one could expect to achieve using a
> solar chimney of sorts in place of large fans in a greenhouse?
>
>
> Thanks for any input,
> Mark
Posted by Anthony Matonak on May 6, 2004, 6:56 am
Mikemotorbike wrote:
> I have read that a solar chimney moves the same as a 1/4 hp motor.
...
Wouldn't it depend greatly on the size and other characteristics
of the solar chimney? Obviously a chimney one foot in diameter
and a couple of feet long is going to move a lot less air than
one as wide as a football field and half a mile tall.
I believe the formulas for calculating this stuff has been posted
before on numerous occasions so a quick Google.com search should
yield enough hits to be useful.
Anthony
Posted by nicksanspam on May 6, 2004, 10:20 am
Say we pump water over 400 ft^2 of roof at night in Phoenix and cover
a ceiling with N 4"x10' 10 ft^2 10 Btu/h-F PVC pipes, in one kind of
wet Cool Cell (tm) system, viewed in a fixed font...
close pipe-air house-outdoor w = 0.0056
for 12h resistance Tin resistance
at night 1/(10N) | 1/200 103.5 F day/
Tw ------/ ---*--www-------*-------www------ 72.9 F night outdoor temp
| | <----- I
| w mass surface resistance
| w 1/3000 (ft^2)
| w
| |
| *--- Tc
| |
--- 55N --- house mass
--- Btu/F --- 6000 Btu/F
| |
- -
How many pipes (N) are needed, if Tin = 80 F max?
The outdoor vapor pressure Pa = 29.921/(1+0.62198/w) = 0.267 "Hg, so Twb
= 9621/(22.47-ln(460+72.9+100Pa-Twb)) = 9621/(22.47-ln(560-Twb)) = 512 R
on the left with 520 R on the right, then 512, 517, 514,... 515 R, ie 55 F.
If 400 ft^2 can keep the water near the 55 F wet bulb temp, and the house air
is 80 F for 12 hours when it's 103.5 F outdoors, I = (103.5-80)200 = 4700
Btu/h and Tc min = 80-4700/3000 = 78.4 F, ignoring the pipes.
The house needs 12x4700 = 56.4K Btu/day of cooling. If the pipes are
55 F at dusk, and the house mass has temp Tc = T, and they store
55N(78.4-55) + 6000(78.4-T) = 56,400 Btu, Tc = 69+0.2145N. If Tc
55+(78.4-55)e^(-12/RCn) = 69+0.2145N, RCn = 12/(4.69-ln(65.3+N)
= 6000(1/3000+1/(10N)) makes N = 300/(6/(4.69-ln(65.3+N))-1).
N = 20 on the right makes N = 12.7 on the left. N = 12.7 on the right
makes N = 17.6, then 14.2, 16.5, 15.0, then 16, so 16 4"x10' pipes
would work, with the help of the thermal mass of the house...
Is it cheaper to evaporate water or use AC for cooling, assuming
suitable weather conditions and 1000 Btu/lb of water and
10,000 Btu/kWh of AC electricity?
At $1.875 for an extra 748 gallons of water ($0.00251/g) and $0.0865/kWh
in the 9 warmest months of the year in Phoenix, water costs $0.00251/(7330)
= 0.34 microcents/Btu, and AC costs 8.65 uc/Btu, so water is 25X cheaper.
Which would Mother Nature prefer? As with embodied energy, it seems
better to follow these present market economics than to try to
outguess the future of state-to-state water agreements.
Nick
Posted by SQLit on May 6, 2004, 2:16 pm
> Say we pump water over 400 ft^2 of roof at night in Phoenix and cover
> a ceiling with N 4"x10' 10 ft^2 10 Btu/h-F PVC pipes, in one kind of
> wet Cool Cell (tm) system, viewed in a fixed font...
> close pipe-air house-outdoor w = 0.0056
> for 12h resistance Tin resistance
> at night 1/(10N) | 1/200 103.5 F day/
> Tw ------/ ---*--www-------*-------www------ 72.9 F night outdoor temp
> | | <----- I
> | w mass surface resistance
> | w 1/3000 (ft^2)
> | w
> | |
> | *--- Tc
> | |
> --- 55N --- house mass
> --- Btu/F --- 6000 Btu/F
> | |
> - -
> How many pipes (N) are needed, if Tin = 80 F max?
> The outdoor vapor pressure Pa = 29.921/(1+0.62198/w) = 0.267 "Hg, so Twb
> = 9621/(22.47-ln(460+72.9+100Pa-Twb)) = 9621/(22.47-ln(560-Twb)) = 512 R
> on the left with 520 R on the right, then 512, 517, 514,... 515 R, ie 55
F.
> If 400 ft^2 can keep the water near the 55 F wet bulb temp, and the house
air
> is 80 F for 12 hours when it's 103.5 F outdoors, I = (103.5-80)200 = 4700
> Btu/h and Tc min = 80-4700/3000 = 78.4 F, ignoring the pipes.
> The house needs 12x4700 = 56.4K Btu/day of cooling. If the pipes are
> 55 F at dusk, and the house mass has temp Tc = T, and they store
> 55N(78.4-55) + 6000(78.4-T) = 56,400 Btu, Tc = 69+0.2145N. If Tc
> 55+(78.4-55)e^(-12/RCn) = 69+0.2145N, RCn = 12/(4.69-ln(65.3+N)
> = 6000(1/3000+1/(10N)) makes N = 300/(6/(4.69-ln(65.3+N))-1).
> N = 20 on the right makes N = 12.7 on the left. N = 12.7 on the right
> makes N = 17.6, then 14.2, 16.5, 15.0, then 16, so 16 4"x10' pipes
> would work, with the help of the thermal mass of the house...
> Is it cheaper to evaporate water or use AC for cooling, assuming
> suitable weather conditions and 1000 Btu/lb of water and
> 10,000 Btu/kWh of AC electricity?
> At $1.875 for an extra 748 gallons of water ($0.00251/g) and $0.0865/kWh
> in the 9 warmest months of the year in Phoenix, water costs
$0.00251/(7330)
> = 0.34 microcents/Btu, and AC costs 8.65 uc/Btu, so water is 25X cheaper.
> Which would Mother Nature prefer? As with embodied energy, it seems
> better to follow these present market economics than to try to
> outguess the future of state-to-state water agreements.
> Nick
Remember the Dentist castle on Camelback Mountain? He had 10,000 gallons
and a fountain. He freely admitted that during 5-6 months of the year it
worked then as the dew point climbs the results were less than satisfactory.
Probably why he never gave tours threw the home during dog days of summer.
Each of the bedrooms had an window shaker.
You will gain some benefit, but an attic fan or a swamp cooler would
probably be a better investment for a short period of time. Evaporation
works well for awhile. I have had both a/c and evap for 35 years in AZ. Now
I have a/c only. I am getting older and the maintenance was killing me. This
system will take a lot of tinkering for very little return. my opinion only
If it works so well why is it not part of every home built in the Phoenix
area. If it worked the utilities would be offering rebates or help to get it
installed so that they could reduce their load.
The cooling medium must be below 50 degrees for humidity removal. We made
an experiment with a chilled water system during a scheduled shut down.
Water was about 65 degrees when we started the chillers. The trouble calls
stopped once the return loop hit 55 degrees. At that point we were producing
enough cold water that the buildings were beginning to work again. Our out
going temp was about 46 degrees. Took over night to satisfy the system
completely.
Please let us know how it works.
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> I am wondering if some of the more technically savvy people out there
> could speak to what kind of air flow one could expect to achieve using a
> solar chimney of sorts in place of large fans in a greenhouse?
>
>
> Thanks for any input,
> Mark