|
Drum Wall Thermal Storage Questions
The house will be built east of Nashville Tennessee
QUESTION:I'm designing a house for my mom (I'm not an architect though), but it's
just a simple house that I'd like to have it heated mostly or entirely by
solar heat. It's a small 24x36 house, 1 bed, 1 bath, and the living room,
dining room and kitchen is all open. I was thinking of using 12'x4' of the
space as a thermal storage closet, using (8) 55 gallon drums of water, and 4
standard sliding glass door replacement glass panels on 4' centers (south
side). She likes it warm, so inside temp will probably be set at 75-80*F
in winter. The house will be built east of Nashville Tennessee. I was
thinking of one set of 6' wide x 6'8" high sliding glass doors (low E) on
the south side), one 36" wide solid pine front door on the west side, and
(4) windows on the north side (approx. 30"w x 24" h each window (low E)). I
was thinking of having sliding pocket panels (insulated) to pull over
windows and possibly sliding glass door at night, and having the walls R-21
and ceiling R-50.
My hopes are to have the solar closet heat the house 100% if possible in
winter (with 5 day reserve) and used as solar chimney in summer to pull
cooler earthtube air into living area and then into top of solar closet and
out roof vent. Maybe have a coil for preheating domestic water before it
goes to the gas water heater? Then for backup heat have a coil array heated
by water heater water and fan blowing air over them to heat the house.
Does this sound possible? Is (8) 55 gallon drums enough? Will I need to
have insulation panel for solar closet glass or will it take in enough heat
to offset the night losses?
ANSWER: Q: I'm designing a house for my mom (I'm not an architect though), but it's
just a simple house that I'd like to have it heated mostly or entirely by
solar heat...
A: Good! The ratio of solar heat to total heat (including backup heat) is
called the solar heating fraction or "solar fraction." Sounds like you
want the solar fraction to be more than 50%, eg 80% or 90%. (One way to
make it 100% is to omit the backup heating system entirely
Q: It's a small 24x36 house, 1 bed, 1 bath, and the living room, dining room
and kitchen is all open...
A: With 8% of the floorspace as R4 windows and 12" R48 SIP walls and 15 cfm
of air leakage (about 1 ACH, and the ASHRAE min standard per person for
fresh air), its thermal conductance might be 72ft^2/R4 = 18 Btu/h-F for
the windows plus 888/48 = 18.5 for the walls plus 864/48 = 18 for the
ceiling plus 15 for the air leaks, a total of about 70 Btu/h-F.
Q: I was thinking of using 12'x4' of the space as a thermal storage closet,
A: About 6% of the 864 ft^2 of floorspace...
Q: using (8) 55 gallon drums of water, and 4 standard sliding glass door
replacement glass panels on 4' centers (south side).
A: You might fit as many as 24 drums in that space (they are about 3' tall and
2' diameter.) With 6' tall panels, you'd have 96 ft^2 of south glazing. Each
drum has about 25 ft^2 of surface. If you heat them with hot air (vs direct
sun), they should have about 10X the glazing surface, ie 960 ft^2, as in
960/25 = 38 drums, or a larger number of smaller containers, eg 4 gallon/
15 liter 9.5"x9.5"x13" stackable plastic tubs with lids made by ROPAK in
Fullerton, CA, which nest nicely for shipping. These are used for packaging
things like powdered detergents and Montmorency red tart dried cherries
from the Cherry Central CO-OP in Traverse City, MI 49684.
Q: She likes it warm, so inside temp will probably be set at 75-80*F
in winter. The house will be built east of Nashville Tennessee.
A: January is the most difficult month for solar heating in Nashville,
according to NREL data, with 1030 Btu/ft^2 of sun that falls on
a south wall on an average 36.2 F day, 730 on a horizontal surface,
and 230, 470, and 480 on north and east and west walls.
The hypothetical house above would need (77.5-36.2)70 = 2891 Btu/h or
69.4K Btu/day to stay 77.5 F in January. If your mom uses, say 300 kWh/mo
of electricity, 34.1K Btu/day might come from that, leaving 35.3K/day
from other sources, or 5x35.3K = 176.4K for 5 cloudy 36.2 F days, like
176.4K/(450(125-85)) = 10 drums cooling from 125 to 85 F.
Q: I was thinking of one set of 6' wide x 6'8" high sliding glass doors
(low E) on the south side),
A: Maybe R4, with 50% solar transmission, ie 40 ft^2 with 10 Btu/h-F,
transmitting 0.5x40x1030 = 20.6K Btu on an average day?
Q: one 36" wide solid pine front door on the west side,
A: Another 3'x7'/R1 = 21 Btu/h-F, collecting 0.0 Btu/day of sun? You might
picture a foam core door instead, darkish outside, with an air gap and
a layer of polycarbonate over that, and 8"x16" vent holes at the top and
bottom, with an automatic foundation vent in the top hole.
Q: (4) windows on the north side (approx. 30"w x 24" h each window (low E)).
A: Maybe 20ft^2/R4 = 5 Btu/h-F, transmitting 0.5x20x230 = 2300 Btu/day.
No windows on the east or west walls?
Q: I was thinking of having sliding pocket panels (insulated) to pull over
windows and possibly sliding glass door at night,
A: Most people stop using things like that, after a while.
Q: and having the walls R-21 and ceiling R-50.
A: The conductance of the house you describe might be 36 Btu/h-F for the
windows and doors plus 879/21 = 42 for the walls plus 864/50 = 17 for
the ceiling plus 30 for the air leaks, about 125 altogether.
It would need (77.5-36.2)125 = 5163 Btu/h or 123.9K Btu/day to stay 77.5 F
in January. With 300 kWh/mo of electrical usage, that leaves 120.5K Btu/day
from other heat sources, or 5x120.5K = 602.4K for 5 cloudy 36.2 F days,
like 602.4K/(450(125-85)) = 33 drums cooling from 125 to 85 F.
Q: My hopes are to have the solar closet heat the house 100% if possible in
winter (with 5 day reserve)...
A: If the 96 ft^2 of closet glazing is say, R2 with 80% solar transmission,
it can collect 0.8x96x1030 = 79.1K Btu on an average day. At 77.5 F, it
might lose 6h(77.5-36.2)96ft^2/R2 = 11.9K, for a net gain of 67.2K. Add
22.9K of sun from windows, and we have 90.1K Btu/day, 25% less than the
120.5K needed to keep the house 77.5 F on an average day, with no reserve
for cloudy days...
Q: Does this sound possible?
A: Not so far. It might be, with more glazing or insulation.
Q: Will I need to have insulation panel for solar closet glass
or will it take in enough heat to offset the night losses?
A: The calcs above assumed an insulated wall behind the closet glazing, with
some holes for air circulation through the closet during the day and little
heat loss through the glazing at night. Removing the wall would raise the
thermal conductance of the house and cloudy day heat storage requirement.
A bubblewall might be nice here. If the sun shines directly on the drums,
they need less surface than air-heated drums, and the "sunspace" can be
warm at night, ie more usable. The thermal mass might be water containers
under a countertop, with 2 layers of polycarbonate on the south side and
insulation on the other 3 sides. If the room is 77.5 F and south sun passes
through a total of 4 layers of R1 polycarbonate with 90% solar transmission
(the 2-layer bubblewall plus 2 thermal mass layers), a square foot of mass
would receive 0.9^4x1030 = 676 Btu/day and lose 24h(T-77.5)1ft^2/R2, which
makes the water temp T = 77.5+676/12 = 134 F on an average day.
Replacing the closet glazing with A ft^2 of bubblewall that's R2 during
the day and R20 at night makes the house conductance 36 Btu/h-F for the
windows and doors plus 17 for the ceiling plus 30 for air leaks plus
(879-A)/21 = for the walls plus 6h/24hA/R2+18h/24hA/R20 for the bubblewall,
on an average day, a total G = 125+0.115A Btu/h-F. Each square foot of
bubblewall collects 0.9x0.9x1030 = 834 Btu per day of sun and loses about
6h(77.5-36.2)1ft^2/R2 = 124, for a net gain of 710. 710A = 24(77.5-36.2)G
makes A = 208 ft^2, eg 28'x8' of the 36' south side of the house, ignoring
the sun that enters the windows and electrical energy usage.
On a cloudy day, with the foam in place, G = 83 for windows and doors and
ceiling plus 655ft^2/R21 = 31 for walls plus 224/20 = 11 for the bubblewall,
a total of 125 Btu/h-F. Over 5 cloudy days, 5x24(77.5-36.2)125 = 619.5K Btu
might come from 619.5K/(134-84) = 12,390 pounds or 1549 gallons or 194 ft^3
of water cooling from 134 to 84 F. This might live under a 4' wide x 3' tall
x 16' long countertop or series of countertops.
The bubblewall might be a well-caulked 28'x8'x10" box with a single layer
of 48" wide Replex polycarbonate plastic on each side of a frame made
from plastic 2x4s with stainless steel screws, with a plastic film liner
at the bottom containing a few inches of detergent solution and a 28'x2"
PVC pipe full of holes connected to a shop vac, with a return at the top
that senses foam vs air and turns off the shop vac when the cavity is full.
|
|
|
|
