Posted by SJC on March 1, 2006, 8:00 pm
If it were only that easy. I believe capitalism works best when =
is readily available. When people start companies on second mortgages
and maxing out credit cards for just enough to go broke, we do not =
many new companies and many new good jobs.
Posted by nicksanspam on March 2, 2006, 2:18 pm
I'm not a builder, but here's a more detailed example: say you wanted to
solar heat a 20' cube in Rochester, NY, where 540 Btu/ft^2 (not much) falls
on a south wall on an average 29 F December day with a 32 F daytime temp...
First, use enough insulation and south sunspace glazing to make a suitable
sunspace air temp (high enough to provide all the heat the cube needs on
an average day with reasonable airflow and low enough to minimize loss
from the glazing to the outdoors). A south sunspace with 20'x20' of U0.58
Thermaglas Plus twinwall polycarbonate (from Palram Americas at 800 994-5626,
about $5 for a 4'x12' sheet) with 80% solar transmission would collect
0.8x400x540 = 172,800 Btu on an average day. With average R20 cube walls
(eg R32 walls and R4 windows, counting air leaks and indoor electrical use),
it might lose 6h(Ts-32)400x0.58 [the daytime glazing loss to the outdoors]
+ 18h(65-29)400/R22 [= 11782 Btu night loss from the south wall]
+ 24h(65-29)4x400/20 [= 69120 Btu from the rest of the walls all day],
which makes the average sunspace air temp Ts = 98 F.
If the "sunspace" is deeper than a few inches, adding a dark mesh curtain
a foot or so from the glazing can make it more comfy for people and lower
the glazing loss by filling the space between the curtain and glazing with
70 F room air.
If 98 F sunspace air heats the cube for 24-hours on an average Dec day (with
a night setback and lots of thermal mass surface inside--this is difficult)
with upper and lower A ft^2 vents and a 20' vent height diff, (11782+69120)/6h
= 13484 Btu/h = 16.6Asqrt(20')(98-70)^1.5 makes vent area A = 1.2 ft^2, with
16.6x1.2xsqrt(20x28) = 482 cfm of natural thermosyphoning airflow.
Next, size the air heater glazing inside the sunspace glazing and find the
air heater temp on an average day. With 160 ft^2 of R1 air heater glazing
with 90% transmission, the sunspace (not the air heater) might collect
0.8x540(400-160) = 103680 Btu/day and lose 6h(98-32)400x0.58 [= 91872 Btu
from the glazing] + 18h(65-29)400/R22 [= 11782 Btu from the south wall at
night] + 24h(65-29)4x400/20 [= 69120 Btu from the rest of the walls all day]
- 6h(Tc-98)160/R1 [the sunspace gain from the air heater glazing], which
makes the average air heater temp Tc = 170 F.
With no heat loss from the cloudy day tank on an average day (a first
approximation) the tank water temp on an average day will be 170 F. Now
find the min water temp needed to keep the cube 70 F. On a cloudy day, the
cube's thermal conductance is 5x20x20/20 = 100 Btu/h-F, so it needs about
(70-29)100 = 4100 Btu/h to stay 70 F during the day (and (60-29)100 = 3100
at night.) If the air heater contains 64' of fin-tube with 5x64 = 320 Btu/h-F
of conductance (behind foamboard insulation with a flow organizer at the top),
the min water temp is about 70+4100/320 = 83 F.
Over 5 cloudy days, the cube needs about 5x24h(65-29)100 = 432K Btu from
432K/(170-83) = 4966 pounds or 620 gallons or 80 ft^3 of water cooling from
170 to 83 F in a 2'x4'x10' EPDM-lined plywood box or a 620 gallon W163 $300
insulated 4' deep x 6' diameter circular tank from STSS at (717) 761-5838,
with one of their $00 HEX WT66-113 120' 3/4" copper pipe heat exchange
coils to preheat water for showers, with the help of two PE pipe coils in 2
55 gallon drums as greywater heat exchangers.
This system could be tweaked more. It's just a bubble diagram of a heating
system, vs a whole house. There are lots of other architectural decisions to
make about shapes and colors and materials, where to put the kitchen sink,
and so on.
Posted by nicksanspam on March 3, 2006, 12:46 pm
If the sunspace air keeps the cube warm for 6 hours, we need to store about
18h(65-29)100 = 64,800 Btu of overnight heat (or less, if the cloudy-day store
supplies some overnight heat, but that increases the tank or air heater size.)
If the cube is 70 F at dusk and 60 at dawn and we store this heat in the
living space in a straightforward way, 60 = 29+(70-29)e^-(RC/18h), so RC
= -18/ln((60-29)/(70-29)) = 64 hours and C = RC/R = RCxG = 6400 Btu/F.
The mass needs lots of surface so the room won't overheat during the day.
Drywall has about 1 Btu/F per board foot, so we might line the 2400 ft^2 cube
interior surface with 2.6" of drywall. A hollow 8x8x16" concrete block has
about 5 Btu/F, so we might make the cube walls with 1800 blocks with the holes
lined up and holes at the top and bottom to allow room airflow, with the R20
insulation on the outside. A 4"x20' PVC pipe holds about 100 pounds of water,
so we might use 64 of them in an 8x8 column. They might pass through some
hollow block holes. Furnishings help too, especially concrete furniture :-)
Mass in direct sun works better, eg a masonry floor or a large aquarium or
a water-filled pedestal near a south window, altho windows in the living space
increase the required size of the cloudy-day store.
But this provides poor room air temp control, and it wastes setback energy:
keeping the cube exactly 70 F for 6 hours and 60 for 18 hours a day requires
less heat (18h(60-29)5x20x20/R20 = 55.8K Btu at "night") than letting it droop
from 70 F at dusk to 60 by dawn. A smaller mass with a larger temp swing can
store the same energy and provide better room air temp control. For example,
if we hang N 4"x20' pipes horizontally under the ceiling and they keep the
cube 60 F at dawn with 3100 Btu/h from 20N ft^2 of surface with 30N Btu/h-F,
Tmin = 60+3100/(30N) = 60+103/N. If the pipes warm to 98 F by dusk and 55.8K
= (98-(60+103/N))100N, N = 18 pipes will do. They could be wrapped with foil
to avoid overheating the room by radiation. As an alternative, we might put
about 1" of water (4.5 psf) in 30" poly film ducts (48" laid flat) on plywood
platforms with foil beneath. We might move warm air down to the room as needed
with a slow ceiling fan and a room temp thermostat and an occupancy sensor.
In this case, a less-massy room will save setback energy.
Shiny ceilings aren't for everyone, but when I asked a doctor about her new
sunspace design with a solar staircase, she said "Shiny ceilings are great!
My favorite architect is Frank Gehry. I have some of his cardboard chairs."
Posted by 3D Peruna on March 3, 2006, 2:35 pm
To each their own... FOG is a "one-trick pony" if you ask me. I think
he's lost sight of what architecture ought to be...just like most of the
modern art/literature/theater world. It's about shock, titillation and
an effort to be different for different sake.
Also, there's little in most of his designs to be "sustainable".
Of course, he'd disagree with my assessment of him, and there are plenty
of people prepared to pay him, and his firm, more money than is
reasonable. Maybe I need to go whacky to become famous.
Posted by nicksanspam on March 4, 2006, 8:08 pm
Here's a 20' cube calc showing the required water depth as a function of
the number of 4'x8' platforms near the ceiling. Shallower water depths
have a larger temp swing, so they store more heat per pound...
20 FOR N=4 TO 8'number of 4'x8' shelves
30 A=N*32'one-sided shelf area (ft^2)
40 T=1'initial water depth (feet)
50 TMIN`+3100/(3*A)'min water temp (F)
60 RCb.33*T/3'time constant (hours)
70 TMAX+(TMIN-98)*EXP(-6/RC)'dusk temp (F)
80 TH5/(A*(TMAX-TMIN))'new water depth (feet)
90 IF ABS(TH-T)>.01 THEN T=TH: GOTO 60'iterate
100 PRINT N;12*T,TMAX,TMIN,62.33*32*T
110 NEXT N
number of water depth dusk water dawn water weight of each =
shelves (inches) temp (F) temp (F) shelf (lb)
4 8.422272 78.16786 68.07291 1399.894
5 3.431941 86.50914 66.45834 570.4343
6 2.308075 90.73237 65.38195 383.6328
7 1.796465 93.14922 64.6131 298.5965
8 1.402662 95.12894 64.03646 233.1412